Equilibrium Expressions & ICE Charts

In order to calculate the equilibrium concentrations of compounds we need to use an ICE Chart. 

Calculating Equilibrium Concentrations

Concept: Understanding an ICE Chart 

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Video Transcript

Hey guys! In this new video, we're going to do something incredibly important – calculating equilibrium concentrations. Up to this point, I've been giving you questions in which our reaction is at equilibrium and the numbers I’m giving you are equilibrium amounts.
Now we’re going to be tasked with figuring out what those equilibrium amounts when I give you just initial concentrations. Remember, initial concentrations and equilibrium concentrations are totally different. We're going to say sometimes we’ll be asked to calculate concentrations at equilibrium after being given initial concentrations. We're going to say in order to do this, we're going to have to use our favorite friend, the ICE chart. Remember, what does ICE stand for? ICE stands for initial, change, equilibrium. We're going to learn how to use an ICE chart and when do we use an Ice chart. What we should realize here is that ICE charts are only allowed to have two types of units. We're going to say they're used to having atmospheres or molarity as the units. Remember, why those two units? Because atmospheres are connected to Kp and molarity is connected to Kc. We're still going to be dealing with our equilibrium constants because they go hand in hand with our ICE charts.
When do we use an ICE chart? You only ask yourself one question or really you say one thing. You're going to say anytime we have more than one compound in our balanced equation without an equilibrium concentration, then we have to use an ICE chart. Before we start this next question, what do I mean by this? In the next question we have three compounds within our balanced equation. Let's say I gave you the equilibrium amounts for two of them. We’d only be missing one person at equilibrium. In that case, we wouldn't need to do an ICE chart. But let's say I gave you only one person at equilibrium.
Technically we’d have to do an ICE chart in order to organize our work to see what the correct answer would be. Again, if you're missing more than one compound at equilibrium, you should use an ICE chart. If you don't have anyone at equilibrium, then you should definitely use an ICE chart.

An ICE Chart should be used when we are missing more than one equilibrium amount for compounds in our balanced equation. 

Since our equilibrium constant K is involved in an ICE Chart then we must continue to ignore solids and liquids. 

Example: We have a solution where Ag(CN)2 (g), CN (g), and Ag+ (g) have an equilibrium constant, K, equal to 1.8 x 10-19. If the equilibrium concentrations of Ag(CN)2 and CN are 0.030 and 0.10 respectively, what is the equilibrium concentration of Ag+?

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Video Transcript

Let's take a look at the first two examples and see how we approach these types of questions. We say here when we have a solution where Ag(CN)2- gas, CN- gas, and Ag+ gas and they have an equilibrium constant K equal to 1.8 times 10 to the negative 19. If the equilibrium concentrations of Ag(CN)2- and CN- are these numbers respectively, what is the equilibrium concentration of Ag+?
We ask ourselves, do we use an ICE chart or not? Here I give you the equilibrium amounts of this compound, which is 0.030 and this compound which is 0.10. You're missing only one equilibrium amount. If you're missing only one equilibrium amount, then you don't have to do an ICE chart. We're simply going to say in this case that K equals products over reactants. We say that products here would be CN- squared times Ag+. Since we're going to need some room guys, I'm going to remove myself from the image so we have more room to work with. Divided by Ag(CN)2-.
What we're going to do now is we're going to plug in the numbers that we know. We know what K is. It's 1.8 times 10 to the negative 19. CN- we said was 0.10 and it's going to be squared. Ag+ is what we're looking for, so we're going to have it as a variable, divided by Ag(CN)2- which is 0.030. This question becomes fairly easy. All we have to do now is isolate our Ag+. What we're going to do first is multiply both sides by 0.030. These two multiply together to give us a new answer of 5.4 times 10 to the negative 21. That equals 0.10 squared times Ag+. We just want Ag+ so we're going to divide it out, the 0.10 squared. Our final answer at the end for silver ion will be 5.4 times 10 to the negative 19 molar.
Again, why didn't we have to use an ICE chart? We didn't have to use an ICE chart because we're missing only one person at equilibrium. When you're missing no one at equilibrium or you're missing only one variable at equilibrium, then you don't need to do an ICE chart. We only do an ICE chart when more than one concentration is missing at equilibrium.

Example: We place 2.5 mol of CO and 2.5 mol of CO3 in a 10.0 L flask and let the system come to equilibrium. What will be the final concentration of CO2?  K = 0.47

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Problem: For the reaction below, KC = 8.3 x 10-10 at 25oC. What is the concentration of N2 gas at equilibrium when the concentration of NO2 is twice the concentration of O2 gas?

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Example: When 0.600 atm of NO­­2 was allowed to come to equilibrium the total pressure was 0.875 atm. Calculate the Kof the reaction. 

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Video Transcript

Hey guys! In this brand-new video, we're going to continue with our conversation on calculating equilibrium concentrations. Let's take a look at this particular question. We're going to say when 0.600 atmospheres of NO2 was allowed to come to equilibrium the total pressure was 0.875 atmospheres. Calculate the Kp of the reaction.
All I'm giving you is 0.600 atmospheres of NO2. I'm saying I'm allowing it to come to equilibrium. What that tells me is this amount is not my equilibrium amount. It's rather my initial amount, so this is my initial pressure. All we say to ourselves is do we have to use an ICE chart? The answer here is yes, because we're going to have this missing an equilibrium amount, this is missing an equilibrium amount and this is missing an equilibrium amount. Anytime more than one of my compounds in my balanced equation is missing an equilibrium amount, I have to do an ICE chart. This number here, I'm giving it time to get to equilibrium which means it's my initial pressure. We're going to say I is initial, C is change, E is equilibrium.
We're going to say initially we're starting out with 0.600 atmospheres. Remember, atmosphere goes with Kp. I don't tell us anything about NO or O2 so they're initially at zero. Remember the important phrase we've been saying, we're losing reactants to make products. This is going to be minus 2x because of the two and it's minus because it's a reactant. Plus 2x, the two there, that's why it’s plus 2x. Plus X bring down everything, 0.600 minus 2x plus 2x plus X.
What we should realize here is they're telling me what the total pressure is. This goes back to Henry's law, or actually not Henry's law but more importantly, Dalton's law. Dalton's law says that the total pressure that we experience is equal to the pressure of all of the gases added up. My total pressure is equal to the pressure of NO2 plus the pressure of NO plus the pressure of O2. At equilibrium, that's my equilibrium total pressure.
What we're going to do here is say at equilibrium, each of these gases is equal to each of these equations. At equilibrium, our total pressure is 0.875 atmospheres. At equilibrium, NO2 is 0.600 minus 2x. At equilibrium, NO is 2x. At equilibrium, O2 is just X. We're going to say that this negative 2 and this positive 2 basically cancel each other out. Our equation becomes 0.875 atmospheres equals 0.600 plus X. We need to isolate X here so we're going to subtract, 0.600, 0.600. X here equals 0.275 atmospheres.
Remember, this is Dalton's law. The total pressure we experience is the pressure from all of the gases added up together. Since this is an equilibrium total pressure, we use the equilibrium equations for each of the compounds. We just isolated what X is, so all we need to do is take this X answer, plug it in here,
plug it in here, plug it in here and we know each one of them at equilibrium. If we know each of them at equilibrium, then it's those equilibrium amounts that we plug in to Kp. Remember, Kp is just products over reactants so it's NO squared times O2 divided by NO2 squared because the two and the two here.
When we plug in the X in for NO2, we get 0.050 atmospheres. When we plug it in for NO, at equilibrium NO equals 2x. Plug in the X that we just found. That's NO. Then O2 is just equal to X at equilibrium so it’s that same exact number. Take all those numbers and plug it in. When you do all that, you'll get your Kp equal to 33.28. To do this question, you had to have a knowledge of Dalton's law. Remember that deals with gases, and then apply that concept to this new idea of an ICE chart. It's kind of taking a little bit of old information combining with some new information in order to obtain our Kp value.

Example: An important reaction in the formation of acid rain listed below.

2 SO2 (g)   +  O2 (g)  ⇌  2 SO3 (g)

Initially, 0.023 M SO2 and 0.015 M O2 are mixed and allowed to react in an evacuated flask at 340 oC. When an equilibrium is established the equilibrium amount of SO3 was found to be 0.00199 M. Calculate the equilibrium constant, Kc, for the reaction at 340 oC. 

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Example: If Kc is 32.7 at 300oC for the reaction below:

H2 (g) +  Br2 (g) ⇌  2 HBr (g) 

What is the concentration of H2 at equilibrium if a 20.0 L flask contains 5.0 mol HBr initially?

8m
Video Transcript

Hey guys, in this brand new video, we're going to continue with our conversation on calculating equilibrium concentrations.
So, let's take a look at this particular question. We're going to say when .600 atmospheres of NO2 was allowed to come to equilibrium the total pressure was .875 atmospheres. Calculate the Kp of the reaction. All I'm giving you is .600 atmospheres of NO2. I'm saying I'm allowing it to come to equilibrium. What that tells me is this amount is not my equilibrium amount. It's rather my initial amount. So, this is my initial pressure. All we say to ourselves is do we have to use an ICE chart? The answer here is yes, because we're going to have this missing an equilibrium amount. This is missing an equilibrium amount and this is missing an equilibrium amount. Anytime more than one of my compounds in my balance equation is missing an equilibrium amount, I have to do an ICE chart. This number here, I'm giving it time to get to equilibrium which means it's my initial pressure.
We're going to say I is initial, C is change, E is equilibrium. We're going to say initially we're starting out with .600 atmospheres. Remember, atmosphere goes with Kp. I don't tell us anything about NO or O2 so they're initially at zero. Remember the important phrase we've been saying. We're losing reactants to make products. This is going to be minus 2x because of the 2. It's minus because it's a reactant. Plus 2x, the 2 there that's why it’s plus 2x. Plus x. Bring down everything. 0600 minus 2x plus 2x plus x.
Now, what we should realize here is they're telling me what the total pressure is. This goes back to Henry’s law, or actually not Henry's law but more importantly Dalton's law. Dalton's law says that the total pressure that we experience is equal to the pressure of all of the gases added up. My total pressure is equal to the pressure of NO2 plus the pressure of NO plus the pressure of O2. At equilibrium, that's my equilibrium total pressure. What we're going to do here is say at equilibrium, each of these gases is equal to each of these equations. At equilibrium, our total pressure is .875 atmospheres. At equilibrium, NO2 is .600 minus 2x. At equilibrium, NO is 2x. At equilibrium, O2 is just x. We're going to say that this negative 2 and this positive 2 basically cancel each other out. Our equation becomes .875 atmospheres equals .600 plus X. We need to isolate x here so we're going to subtract .600, .600. x here equals .275 atmospheres.
Remember, this is Dalton's law. The total pressure we experience is the pressure from all of the gases added up together. Since this is an equilibrium total pressure, we use the equilibrium equations for each of the compounds. We just isolated what x is, so all we need to do is take this x answer, plug it in here, plug it in here, plug it in here and we know each of them at equilibrium. If we know each of them at equilibrium, then it's those equilibrium amounts that we plug in to Kp. Remember Kp is just products over reactants, so NO squared times O2. Divide it by NO2 squared because the 2 and the 2 here. So, when we plug in the x in for NO2, we get .050 atmospheres. When we plug it in for NO, at equilibrium NO equals 2x. Plug in the x so we just found, that's NO. Then is just equal to x at equilibrium so it’s that same exact number. Take all of those numbers and plug it in. When you do all that, you'll get your Kp equal to 33.28.
To do this question, you had to have a knowledge of Dalton's law. Remember, that deals with gases and then apply that concept to this new idea of an ICE chart. It's kind of taking a little bit of old information, combining with some new information in order to obtain our Kp value. 

Problem: At a given temperature the gas phase reaction:

H2 (g) +  O2 (g) ⇌ 2 NO (g)

has an equilibrium constant of 4.00 x 10-15. What will be the concentration of NO at equilibrium if 2.00 moles of nitrogen and 6.00 moles oxygen are allowed to come to equilibrium in a 2.0 L flask.

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Example: Consider the following reaction:

COBr(g)  ⇌  CO (g) + Br(g)

A reaction mixture initially contains 0.15 M COBr2. Determine the equilibrium concentration of CO if Kc for the reaction at this temperature is 2.15 x 10-3.

8m

Equilibrium Expressions & ICE Charts Additional Practice Problems

The equilibrium constant Kc for the reaction

H2(g) + I2(g) ⇌ 2H1(g)

is 54.3 at 430 C. At the start of the reaction, there are 0.714 mole of H_2, 0.984 mole of I2, and 0.886 mole of HI in a 2.40 L reaction chamber. Calculate the concentrations of the gases at equilibrium.

Watch Solution

A 0.682- g sample of ICl(g) is placed in a 625 - mL reaction vessel at 682 K. When equilibrium is reached between the ICl(g) and I2 (g) and Cl2(g) formed by its dissociation, 0.0383 g I2 (g) is present. What is KC for the reaction?



Watch Solution

Is a mixture of 0.0205 mol NO2(g) and 0.750 mol N2O4(g) and 0.750 mol N2O4(g) in a 5.25 - L flask at 25 ° C at equilibrium ? If not, in which direction will the reaction proceed - toward products or reactants?
N2O4(g)⇌ 2NO2(g) KC = 4.61x10-3 at 25 °C
a. The reaction will proceed to the right, toward products
b. The reaction will proceed to the left, toward reactants.
c. The mixture is at equilibrium.



Watch Solution

For the dissociation of I2 (g) at about 1200°C, I2 (g) ⇌ 2 I (g), Kc + 1.1 x10-2. What volume flask should we use if we want 0.42 mol I to be present for every 1.00 mol I2 at equilibrium? Express your answer using two significant figures.


Watch Solution

If Kc = 0.0055 for the reaction

N2 (g) + 3 H2 (g) 2 NH3 (g)

runs at 355.0 K, what is Kc at 355.0 K for the reaction:

1/3 N2 (g) + H2 (g) 2/3 NH3 (g)


Watch Solution

The equilibrium constant Kc for the reaction:

C ⇌ D + E 

is 8.10 x10-5. The initial composition of the reaction mixture is [C} = [D] = [E] =   1.10x10^3 M. What is the equilibrium concentration of C, D, and E?


Watch Solution

A key step in the extraction of iron from its ore is FeO(s) + CO (g) Fe (s) + CO_2 (g) K_p = 0.403 at 1, 000 degree C This step occurs in the 700 degree C to 1, 200 degree C zone within a blast furnace. What are the equilibrium partial pressures of COg) and CO_2 (g) when 1.2200 atm of CO (g) and excess FeO(s) react in a sealed container at 1, 000 degree C?



Watch Solution

Enter your answer in the provided box. For the reaction H2(g) + CO2(g) <-->H2O(g) + CO(g) at 700C Kc = 0.534. Calculate the number of moles of H_2 that are present at equilibrium if a mixture of 0.650 mole of CO and 0.650 mole of H_2O is healed to 700 C in a 30.0-L container. 



Watch Solution

Gaseous ammonia was introduced into a sealed container and heated to a certain temperature: 2 NH3(g) N2(g) + 3 H2(g) At equilibrium, [NH3] = 0.0237 M, [N2] = 0.115 M, and [H2] = 0.345 M. Calculate Kc for the reaction at this temperature.

Watch Solution

The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal:

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)

At a given temperature,Kp= 2.7. If 0.13 mol of CO, 0.56 mol of H2O, 0.62 mol of CO2, and 0.43 mol of H2 are put in a 2.0-L flask, in which direction does the reaction proceed ?

Watch Solution

A Student ran the following reaction in the laboratory at 502 K 

PCl3(g) + Cl2(g) PCl_5(g)
When she introduced 0.199 moles of PCI3(g) and 0.235 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 6.84 times 10^-2 M. Calculate the equilibrium constant, Kc. she obtained for this reaction.

Watch Solution

The equilibrium constant, Kc, for the following reaction is 7.00 x 10^-5 at 673 K.
NH4I(s) NH3(g) + HI(g)
Calculate the equilibrium concentration of HI when 0.411 moles of NH4I(s) are introduced into a 1.00 L vessel at 673 K.

Watch Solution

Given the two reactions:

H2S <--> HS^- + H^+, K1 = 9.56 x 10^-8 and 

HS^- <---> S^2- + H^+, K2 = 1.44 x 10^-19.

What is the equilibrium constant Kfinal for the following reaction?

S^2- + 2H^+ <-->H2S

Watch Solution

Nitrosyl bromide, NOBr, is formed in the reaction of nitric oxide, NO, with bromine, Br2 

2NO(g) + Br2(g) --> 2NOBr(g) 

The reaction rapidly establishes equilibrium when the reactants are mixed. At a certain temperature the initial concentration of NO was 0.400 M and that of Br2 was 0.265 M. At equilibrium the concentration of NOBr was found to be 0.250 M. What is the value of Kc at this temperature. Express your answer numerically.

Watch Solution

Consider the reaction:
1/2 N2(g) + O2(g) <--> NO2(g)
Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb for reactions a and b below:

a) N2O4(g) <--> 2NO2(g) Ka

b) N2(g) + 2O2(g) <--> N2O4(g) doubleheadarrow N2O4(g) Kb

Watch Solution

For the reaction shown here:
2H2O(g)  ⇌ 2H2(g)+O2(g)

the equilibrium concentrations were found to be:

[H2O]=0.250 M
[H2]=0.560 M
[O2]=0.800 M

What is the equilibrium constant for this reaction?
Keq =?

 

 

Watch Solution

Consider the following reaction:

2NO(g) + Br2(g) ⇌ 2NOBr(g)
Kp= 28.4 at 298 K

In a reaction mixture at equilibrium, the partial pressure of NO is 107 torr and that of Br2 is 166 torr.

What is the partial pressure of NOBr in this mixture?

 

Watch Solution

For the reaction, 

CO(g) + Cl2(g) ⇌ COCl2(g)       Kc = 255 at 1000 K. 

If a reaction mixture initially contains a CO concentration of 0.1550 M and a Cl2 concentration of 0.171 M at 1000 K.

What is the equilibrium concentration of CO at 1000 K?

What is the equilibrium concentration of Cl 2 at 1000 K?

What is the equilibrium concentration of COCl2 at 1000 K?

Watch Solution

Consider the reaction:

2NO(g)+Br2(g)⇌2NOBr(gKp=28.4 at 298 K

In a reaction mixture at equilibrium, the partial pressure of NO is 126 torr and that of Br2 is 159torr .

What is the partial pressure of NOBr in this mixture?

Watch Solution

At 850 K the equilibrium constant for the following reaction is K c = 15.

2SO2(g) + O2(g) ⇌ 2SO3(g)
 

If we mix the following concentrations of the three gases, predict in which direction the reaction will proceed toward equilibrium: Left, Right, or No Reaction.

a. [SO2] = 0.60 M, [O2] = 0.70 M, [SO3] = 0.16 M

b. [SO2] = 0.20 M, [O2] = 0.60 M, [SO3] = 0.60 M

c. [SO2] = 0.10 M, [O2] = 0.20 M, [SO3] = 0.40 M

Watch Solution

Consider the following reaction:

SO2Cl2(g) ⇌ SO2(g) + Cl2(g)

Kc = 2.99 × 10−7 at 227C

If a reaction mixture initially contains 0.177 M SO2Cl2, what is the equilibrium concentration of Cl 2 at 227C?

Watch Solution

For the reversible reaction

A(g) ⇌ B(g)

which k values would indicate that there is more B than A at equilibrium?

i. K = 9 x 10-9

ii. K = 0.2

iii. K = 9000

iv. K = 9 x 109

Watch Solution

The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles in liquid water according to the equation:
2HgO(s) + H2O(l) + 2Cl2(g) ⇌ 2HOCl(aq) + HgO(s) + HgCl 2(s)


What is the equilibrium constant expression for this reaction?

Watch Solution

What is the proper form of the equilibrium constant expression for the equation: N  2(g) + O2(g) ⇌ 2NO(g)?

Watch Solution

What is the value of Q when the solution contains 2.50x10M Mg2+ and 2.00x10M CO32?

Watch Solution

Calculate the equilibrium constant for each of the reactions at 25°C.

a. 2Fe+3(aq) + 3Sn(s) → 2Fe(s) + 3Sn +2(aq)

b. O2(g) + 2H2O(l) + 2Cu(s) → 4OH-(aq) + 2Cu+2(aq)

c. Br2(l) + 2I-(aq) → 2Br-(aq) + I2(s)

Watch Solution

The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles in liquid water according to the equation

2HgO(s) + H2O(l) + 2Cl2(g) ⇌ 2HOCl(aq) + HgO⋅HgCl 2(s)

What is the equilibrium-constant expression for this reaction?

a. K = [HOCl]/[Cl2] 2

b. K = [HOCl] 2[HgO⋅HgCl2] / [Cl2]2[H2O][HgO]2

c. K = [HOCl]2 / [Cl2]2

d. K = [HOCl]2 [Cl2]2 / [H2O]

Watch Solution

The equilibrium-constant of the reaction

NO2(g) + NO3(g) ⇌ N2O5(g)

is K = 2.1 × 10−20. What can be said about this reaction?

a. At equilibrium the concentration of products and reactants is about the same.

b. At equilibrium the concentration of products is much greater than the concentration of reactants.

c. At equilibrium the concentration of reactants is much greater than that of products.

d. There are no reactants left over once the reaction reaches equilibrium.

Watch Solution

Identify the proper form of the equilibrium-constant expression for the equation

N2(g) + O2(g) ⇌ 2NO(g)

a. = [NO] / [N2][O2]

b. = [NO]/ [N2][O2]

c. = [N2][O2] / [NO]2

d. = 2[NO] / [N2][O2]

Watch Solution

At a certain temperature, the equilibrium constant,  Kc for this reaction is 53.3.

H2(g) + I2(g) ⇌ 2HI(g)

At this temperature, 0.300 mol of H2 and 0.300 mol of I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

 

Watch Solution

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.

H2(g) + I2(g) ⇌ 2HI(g)

At this temperature, 0.400 mol of H2 and 0.400 mol of I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

Watch Solution

The equation for the formation of hydrogen iodide from H 2 and I2 is:

H2(g) + I2(g) ⇌ 2HI(g)

The value of Kp for the reaction is 71 at 710.0 °C. What is the equilibrium partial pressure of HI in a sealed reaction vessel at 710.0 °C if the initial partial pressures of H2 and I2 are both 0.100 atm and initially there is no HI present?

Watch Solution

Phosphorus pentachloride decomposes according to the chemical equation.

PCl5(g) ⇌ PCI3(g) + Cl2(g)             Kc = 1.80 at 250 °C

A 0.250 mol sample of PCl5(g) is injected into an empty 2.50 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Watch Solution

For the reversible reaction

A(g) ⇌ B(g)

which k values would indicate that there is more B than A at equilibrium?

 

Watch Solution

A + 2B → 2C      Kc = 2.39

2C → D               Kc = 0.176

Calculate the value of the equilibrium constant for the reaction

D → A + 2B         Kc = ?

Watch Solution

Calculate the value of the equilibrium constant for the reaction: D in equilibrium with A+2B.

A+2B in equilibrium with 2C Kc= 2.35

2C in equilibrium with D Kc= 0.182

Watch Solution

What is the value of K for this aqueous reaction at 298 K?

A+B ⇌ C+D          ΔG° = 10.33 kJ/mol

Watch Solution

Calculate the concentrations of H2SO4. HSO4-. SO42-. and H+ ions in a 0.14 M sulfuric acid solution at 25°C. Assume H+ ions and H3O+ ions to be the same ions. (Ka2 for sulfuric acid is 1.3 x 10-2.)

(H2SO4) = ____

(HSO4-) = _____

(SO42-) = ______

(H-+) = ______

Watch Solution

Calculate the percent ionization of hypochlorous acid (HClO) in solutions of each of the following concentrations (Ka = 3.0 x 10 -8.)

i) 0.230 M

ii) 0.576 M

iii) 0.824 M

Watch Solution

Calculate the value of the equilibrium constant for the reaction:

D in equilibrium with A + 2 B, A + 2 B in equilibrium with 2 C, K  = 2.35, 2C in equilibrium with D, K= 0.182

Watch Solution

An equilibrium mixture of H2, I2 and HI at 458°C contains 1.34 atm of H2, 1.34 atm of I2 and 9.30 atm of HI.

What is the equilibrium constant (Kp) for this reaction?

a) 31.6

b) 48.2

c) 64.7

d) 78.3

e) 92.5

Watch Solution

Considering the following equilibrium, what is the expression for K?

a) [HI]2/[H2]
b) [H2][I2]/[HI]
c) ([H2][I2])1/2 /[HI]2
d) [HI]/ [H2]1/2
e) [HI]2/[H2][I2]

Watch Solution

An equilibrium constant with a small magnitude indicates that a system favors _____ when it reaches equilibrium.

a) reactants

b) products

c) neither reactants nor products

d) both reactants and products

e) none of the above

Watch Solution

At a certain temperature, the Kc of the reaction below is 1.5 x10 −11. If 0.10 mol N2 and 0.10 mol O2 are reacted together in a 1.00 L container, how much NO is present at equilibrium?
N2 (g) + O2 (g) ⇌ 2 NO (g) Kc = 1.5x10−11

A. 1.5x10−13 M

B. 0.20 M

C. 3.9x10−5 M

D. 1.9x10−5 M

Watch Solution

When excess PbCl2 is dissolved in 1.00 L of water, 0.032 mol of Cl  forms at equilibrium. What is the Kc of the reaction:

PbCl2 (s) ⇌ Pb2+ (aq) + 2 Cl  (aq) Kc = ?

A. 1.0x10−3

B. 1.3x10−4

C. 3.3x10−5

D. 1.6x10−5

Watch Solution

At a certain temperature, 0.91 mol of NO is placed in a 1.0 L vessel. Once the equilibrium is established, 0.22 mol of each product is present. What is the Kc of the reaction?

2 NO (g) ⇌ N2 (g) + O2 (g)

A. 0.22

B. 0.81

C. 1.2

D. 1.0

Watch Solution

For the reaction 2 SO3 (g) ⇌ 2 SO2 (g) + O2 (g), which of the following is the correct equilibrium expression for Kc?

A. Kc = [SO3]2 / [SO2]2 [O2]

B. Kc = [SO2]2 [O2] / [SO3]2

C.Kc = 2 [SO2] [O2] / 2 [SO3]

D. K= 2 [SO3] / 2 [SO2] [O2]

Watch Solution

Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation:

PCl5(g) ⇌ PCl3(g) + Cl2(g)

At 250° 0.125 M PCl5 is added to the flask. If Kc = 1.80, what are the equilibrium concentrations of each gas?

Watch Solution

Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows:

[N2]eq = 3.6 M, [O2]eq = 4.1 M, [N2O]eq = 3.3 × 10-18 M.

2 N2(g) + O2(g) ⇌ 2 N2O(g)

Watch Solution

Which is the correct equilibrium constant expression for the following reaction?

2Fe (s) + 3H2O (g) ⇌ Fe2O3 (s) + 3H2 (g)

Watch Solution

Calculate the pH of a 0.250 M NH 4F aqueous solution. The Kb (NH3) = 1.76 × 10 -5 Ka (HF) = 3.5 × 10-4.

Watch Solution

Calculate the pH of a 0.0050 M HF aqueous solution. The Ka for HF is 3.5 × 10  - 4

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Calculate the pH of a 0.100 M HClO2 aqueous solution. The Ka for HClO2 is 1.1 × 10-2

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Find the [OH-] and pH of a 0.33 M methylamine (CH 3NH2) aqueous solution.

Kb (CH3NH2) = 4.4 × 10-4.

Watch Solution

Calculate the pH for a 0.20 M pyridine (C 5H5N) solution. Kb = 1.7 × 10-9

Watch Solution

Morphine is a weak base. A 0.150 M solution of morphine has a pH of 10.5. What is the Kb for morphine.

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Calculate the pH of an aqueous 0.10 M NH4Cl solution. The Kb for NH3 (ammonia) is 1.76 × 10-5.

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Calculate the [H3O+] and the pH of a 0.80 M aqueous HClO2 solution. The Ka for HClO2 is 0.011.

Watch Solution

A hypothetical reaction is shown below. 2A(g) +B2(g) ↔ 2C(g). A flask of 1 L volume is charged with 0.382 mol A and 0.0952 mol of B2 and allowed to come to equilibrium. At equilibrium, the vessel is found to contain 0.0624 mol of C. What is the value of Kc for this reaction?

A) 0.190

B. 1.79

C. 0.00193

D. 2.46

E. 0.596

Watch Solution

Consider the following reaction at a certain temperature. 

2H2O (g) ⇌ 2H (g) + O2 (g)

An equilibrium mixture of the reaction at that same temperature was found to contain 0.416 M of water vapor, H2O, 0.398 M oxygen gas, O2, and 0.472 M hydrogen gas, H 2. Determine the value of the equilibrium constant, Kc, for this reaction.

A. 1.09

B. 0.452

C. 0.512

D. 0.432

E. 0.309

Watch Solution

The equilibrium constant, Kc, for the reaction. 2A(g) + 3B(g) → 3C(g) is 2.85 x 10  3 at 48°C. What is the value of Kp at this temperature?

A. 1.98 x 106

B. 184

C. 4.10

D. 4.43 x 104

E. 198 x 106

Watch Solution

Select the correct equilibrium expression for the following reaction.

NaH(s) + H2O(l) → NaOH(aq) + H2(g)

Watch Solution

Kc = 1.03 x 10-6 for the reaction shown below. if the equilibrium concentrations of H 2 = 0.00463 M and S2 = 0.00684, what is the equilibrium concentration of the H 2S?

2H2S(g) ↔ 2H2(g) +S2(g)

A. 0.142 M

B. 7.02 M

C. 2.35 M

D. 0.377 M

E. 5.54 M

Watch Solution

The equilibrium constant Kc = 6.35 for the reaction shown below. The enthalpy change for this reaction at 25°C is -25.6 kJ.

2 A + B → 3C

What is the correct value of K for the reaction shown below?

3/2 C → A + 1/2 B

A. 0.397

B. 0.079

C. 2.52

D. 0.157

E. 3.18

Watch Solution

Choose the answer which gives the equilibrium expression for the aqueous reaction shown below.

Si(OCH3)4(aq) + 2 H2O(l) ⇌ SiO2(s) + 4 HOCH3(aq)

Watch Solution

Choose the answer which gives the correct equilibrium constant expression for the reaction shown below. 

6 H2(g) + P4(g) ⇌ 4 PH3(g)

Watch Solution

The expression for K for the reaction

4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H 2O(g)

at equilibrium is:

Watch Solution

Now let’s consider the overall reaction as an equilibrium process, carried out at a lower temperature such that H 2O is obtained as a liquid.

2NO(g) + H 2(g) ⇌ N 2O(g) + H 2O(l)

Write the equilibrium expression for this reaction.

 

Watch Solution

Consider the following reaction and its equilibrium constant:

                FeO(s) + CO(g) ⇌ Fe(s) + CO2(g)          Keq = 0.67

        If the equilibrium concentration of CO is measured at 0.40 M, what is the equilibrium concentration of CO 2?

        A)  0.40 M    B)  1.1 M    C)  0.67 M    D)  0.27 M    E)  1.7 M

 

 

Watch Solution

Write the equilibrium constant expression for the reaction:  2HgO( s) ⇌ 2Hg(l) + O 2(g)

 

 

Watch Solution

Consider the reaction shown below with its equilibrium constant:

            S 2Cl 2(g) + Cl 2(g) --->  2SCl 2(g)  Keq = 4

Examine the figure, and determine if the system is at equilibrium. If it is not, in which direction will it proceed to reach equilibrium?

    A)    The reaction is at equilibrium.

    B)    The reaction is not at equilibrium, it will shift to the left.

    C)    The reaction is not at equilibrium, it will shift to the right.

    D)    It is not possible to tell if the reaction is at equilibrium.

    E)    To reach equilibrium, the value of Keq must change

Watch Solution

Which equilibrium constant represents a reaction that is most product favored?

    A) Keq = 0.025      B) Keq = 5.2        C) Keq = 8.4 × 10−5        D) Keq = 6.3 × 105

 

Watch Solution

For the reaction,

CoO(s) + H 2(g) ⇌ Co(s) + H 2O(g)

at 550°C, the value of K is 67.  The equilibrium constant expression is

 

Watch Solution

Which of the following is the correct K b expression for the reaction below:

B(aq) + H2O(l) ⇌ HB+ (aq) + OH (aq)

Watch Solution

Find the pH of a 0.135 M aqueous solution of HCIO 2 , for which Ka = 1.1 x 10–2 .

A) 1.25

B) 3.28

C) 1.17

D) 1.34

E) 1.48

Watch Solution

Determine the pH of a 0.15 M aqueous solution of KF. For HF,  Ka = 7.0 × 10−4.

A) 5.85            B) 12.01          C) 8.17            D) 6.68            E) 2.32

Watch Solution

Determine the pH of a 0.35 M aqueous solution of CH 3NH 2 (methylamine). The b of methylamine is 4.4 × 10 −4.     

A) 13.24      B) 3.86      C) 10.00      D) 12.09      E) 1.96

Watch Solution

The Ka of hydrazoic acid (HN3) is 1.9 × 10 −5 at 25.0°C. What is the pH of a 0.35 M aqueous solution of HN 3?

A) 5.23      B) 2.59      C) 2.41      D) -2.46      E) 1.14

Watch Solution

The research and development unit of a chemical company is studying the reaction of CH4 and H2S, two components of natural gas:

CH4(g) + 2H2S(g) ⇌ CS2(g) + 4H2(g)

In one experiment, 1.00 mol of CH4, 1.00 mol of CS2, 2.00 mol of H2S, and 2.00 mol of H2 are mixed in a 250-mL vessel at 960 oC. At this temperature, Kc = 0.036. If [CH4] = 5.56 M at equilibrium, what are the equilibrium concentrations of H2?

 

A) 5.56 M                   B) 1.76 M                    C) 6.44 M                    D) 2.44 M

Watch Solution

One reaction that occurs in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and carbon dioxide. The equilibrium constant, p, for the reaction at 1000 K is 0.259.

FeO (s) + CO (g) ⇌ Fe (s) + CO (g)                 p = 0.259 at 1000 K.

What is the equilibrium partial pressure of CO at 1000 K if the initial partial pressures are CO = 1.000 atm and CO2 = 0.500 atm?

A) 1.191                B) 0.121          C) 0.241          D) 0.359          E) 0.191

Watch Solution

For the reaction: H (g) + Br (g) ⇌ 2 HBr (g), c = 7.5 × 10 2 at a certain temperature.

If 2 mole each of H 2 and Br 2 are placed in 2-L flask, what is the concentration of H 2 at equilibrium?  

A) 0.96      B) 0.93            C) 1.86            D) 0.04            E) 0.07

Watch Solution

Given the equilibrium constant and initial concentrations in a reaction, predict the direction the reaction will proceed

 

The equilibrium constant is equal to 5.00 at 1300 K for the reaction:

2 SO (g) + O (g) ⇌ 2 SO (g). If initial concentrations are [SO 2] = 1.20 M, [O 2] = 0.45 M, and [SO 3] = 1.80 M, the system is

A) not at equilibrium and will remain in an unequilibrated state.

B) at equilibrium.

C) not at equilibrium and will shift to the left to achieve an equilibrium state.

D) not at equilibrium and wil shift to the right to achieve an equilibrium state

Watch Solution

The following pictures represent mixtures of A 2B 4 molecules and AB 2 molecules, which interconvert according to the equation:  A 24 ⇌ 2 AB 2.

If mixture (1) is at equilibrium, which of the other mixtures are also at equilibrium?

A) mixture (2)

B) mixture (3)

C) mixture (4)

D) None of the other mixtures are at equilibrium

Watch Solution

1.75 moles of H2O2 were placed in a 2.50 L reaction chamber at 307ºC.  After equilibrium was reached, 1.20 moles of H2O2 remained.  Calculate the equilibrium constant, Kc, for the reaction

2 H2O2(g) ⇋ 2 H2O(g) + O2(g).

A)  2.0 x 10–4   

B)  2.3 x 10–2   

C)  2.4 x 10–3   

D)  5.5 x 10–3   

E)  3.9 x 10–4

Watch Solution

Write the equilibrium expression, Kc, for the following chemical reaction equation.

2 C6H6(g) + 15 O2(g) ⇋ 12 CO2(g) + 6 H2O(g)

Watch Solution

What is the equilibrium equation for the following reaction?

C 2 H 4 (g) + 3 O 2 (g) ⇌ 2 CO 2 (g) + 2 H 2 O (l)

 

Watch Solution

Suppose we put 1.0 mol of HI(g), 1.0 mol of H (g), and 1 mol of I(g) in a 2.0 liter reaction vessel and the following equilibrium is established:
2 HI (g) ⇌ H(g) + I(g)
If Kc = 10 for this reaction at the temperature of the equilibrium mixture, compute the equilibrium concentration of HI.

1. 0.260 M

2. 0.205 M

3. 0.295 M

4. 0.240 M

5. 0.102 M

6. 0.429 M

7. 0.145 M

8. 0.071 M

Watch Solution

Consider the reaction 
C (s) + CO2 (g) → 2 CO(g) .
At equilibrium at a certain temperature, the partial pressures of CO(g) and CO2(g) are 1.22 atm and 0.780 atm, respectively. Calculate the value of K for this reaction.

1. 1.56

2. 1.91

3. 3.13

4. 0.640

5. 2.00

Watch Solution

The  equilibrium constant for the reaction 

                              2 H   2 (g)  +   CO (g) ⇌ CH 3OH (g) 

is 1.6 x 10-8 at a certain temperature. If there are 1.17 moles of H 2 and 3.46 moles of CH3OH at equilibrium in a 5.60 L flask, how many moles of CO are present at equilibrium? 

Watch Solution

One of the most important industrial sources of ethanol is the reaction of steam with ethane derived from crude oil. The reaction is represented by the following thermochemical equation:

Watch Solution

Given the following information,

HF(aq) ⇌ H+ (aq) + F - (aq)                      K  c = 6.8 X 10-4

H2C2O4 (aq) ⇌ 2H+ (aq) + C 2O4-2 (aq)      K c = 3.8 X 10-6

determine the value of Kc for the reaction

2HF(aq) + C2O4-2 (aq) ⇌ 2F - (aq) + H2C2O4 (aq)

A) 0.12

B) 1.2x10-3

C) 1.20

D) 3.50

E) 2.3 x 10-4

Watch Solution

Express the equilibrium constant for the following reaction.

P4O10 (s) ⇌ P(s) + 5 O(g)

A) K = [P4][O2]5 / [P4O10]

B) K = [P4O10] / [P4][O2]5

C) K = [O2-5

D) K = [O2]5

E) K = [P4O10] / [P4][O2]1/5

Watch Solution

Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows:

[N2]eq = 3.6 M, [O2]eq = 4.1 M, [N2O]eq = 3.3 x 10-18 M.

2N2(g) + O2(g) ⇌ 2N2O(g)

A) 2.2 x 10-19

B) 4.5 x 1018

C) 2.0 x 10-37

D) 5.0 x 1036

E) 4.9 x 10-17

Watch Solution

Consider the following reaction at 388 K

2 SO2 (g) + O2 (g) ↔ 2 SO3 (g)

At equilibrium the reaction mixture contains 1.28 atm of O 2 and 6.78 atm of SO3. The equilibrium constant, Kp, at this temperature is 46.7. Calculate the equilibrium partial pressure of SO.

a. 1.14 atm

b. 0.877 atm

c. 0.769 atm

d. 1.30 atm

e. 1.12 atm

Watch Solution

Which is the correct equilibrium constant expression for the following reaction?

Fe2O3(s) + 3 H2 (g) <==> 2 Fe (s) + 3 H2O (g)

a. Kc = [Fe2O3][H2] / [Fe] 2[H2O]3

b. Kc = [H2] / [H2O]

c. Kc = [H2O]3 / [H2]3

d. Kc = [Fe]2[H2O]3 / [Fe2O3][H2]3

e. Kc = [H2O]3 / [H2]3

Watch Solution

What is the percent ionization of 0.50 M NH 3 (aq) solution? The Kb of NH3 is         1.8x10–5

a) 0.50 %

b) 0.60 %

c) 0.70 %

d) 0.80 %

e) 0.90 %

Watch Solution

Which of the statements is FALSE regarding the equilibrium constant, K c?

a) If Kc is greater than one the equilibrium is said to lie towards products.

b) Kc for the B ⇌ A reaction is the inverse of the K c for the A ⇌ B reaction.

c) The value of Kc depends on the physical states of the reactants.

d) Kc is constant regardless of the temperature of the reaction.

e) Kc is a unitless constant.

Watch Solution

Consider this reaction.

2 C(s) + O(g) ⇌ 2 CO (g)

What is the equilibrium expression for this reaction?

 

 

Watch Solution

Consider this reaction.

2 SO(g) ⇌ 2 SO(g) + O(g)

What is the correct Kp expression for this reaction?

 

Watch Solution

Consider this reaction.

P(s) + 6 Cl(g) ⇌ 4 PCl(g)

What is the correct Kc expression for this reaction?

 

 

Watch Solution

Write the equilibrium constant for the following thermal decomposition reaction:

2NaHCO3 (s) ⟺ Na2CO3 (s) + CO2 (g) + H2O (g)

Watch Solution

For the equilibrium,

SO2(g) + NO2(g) ⇌ SO3(g) + NO(g)

The four gases are mixed in a container in the following partial pressures (in atmospheres): SO 2 (2.3); NO2 (2.8); SO3 (1.4); and NO (3.4). At equilibrium it was found that the SO partial pressure was 2.8 atmospheres, what is Kp?

a) 0.28

b) 1.79

c) 2.87

d) 0.134

e) 3.54

Watch Solution

Methylamine is a weak base in water according to the equilibrium

CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH(aq)

What is the correct expression for the equilibrium constant?

Watch Solution

A mixture of 1.00 atm of NO, 0.50 atm of H 2, and 1.00 atm of N2 was allowed to reach equilibrium according to the reaction given below (initially there was no H2O). At equilibrium, the partial pressure of NO was found to be 0.62 atm. Determine the value of the equilibrium constant, Kp, for the reaction:

2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g)

a) 6.08

b) 31.0

c) 215.0

d) 26.5

e) 651.0

Watch Solution

Carbon monoxide gas reacts with hydrogen gas at elevated temperatures to form methanol according to this equation: CO (g) + 2 H(g) ⇌ CH3OH (g). When 0.4 moles of CO and 0.30 moles of hydrogen gas are allowed to reach equilibrium in a 1.0 L container, 0.05 moles of methanol are formed. What is the value of Kc?

  1. 0.50
  2. 3.57
  3. 1.7
  4. 5.4
Watch Solution

When glucose, a sugar, reacts fully with oxygen,carbon dioxide and water are produced according to the equation:

C6H12O6(s) + 6 O2(g) ⟺ 6 CO2(g) + 6 H2O(l)

Write the expression for Kc for this reaction.

Watch Solution

When heated at high temperatures, a diatomic vapor dissociates as follows:

A2 (g) ⟺ 2A (g)

In one experiment, a chemist finds that when 0.0520 mole A 2 was placed in a flask of volume 0.527 L at 590 K, the fraction of A2, dissociated was 0.0238.

 

a) Calculate Kc for the reaction at this temperature

 

 

 

 

 

 

 

 

 

b) Calculate Kp for the reaction at this temperature

 

 

Watch Solution

Which one of the following will change the value of an equilibrium constant?

a) changing temperature

b) adding other substances that do not react with any of the species involved in the equilibrium

c) varying the initial concentrations of reactants

d) varying the initial concentrations of products

e) all of these

Watch Solution

For the following chemical reactions:

Original: 2 H2S (g)⇌ 2 H2 (g) + S2 (g)                   K  P = 2.4 X 10  −4

New : 2 H2 (g) + S2 (g) ⇌ 2 H2S (g)                       K  P’ = ????

Solve for KP’ of the new chemical reaction and select the best description below.

 

KP’                          Original Reaction                  New Reaction

a. 4.2 X 103            Favors products                    Favors reactants

b. 4.2 X 103            Favors reactants                   Favors products

c. −2.4 X 10−4        Favors reactants                   Favors reactants

d. −2.4 X 10−4        Favors products                    Favors products  

e. 2.4 X 10−4          Favors reactants                    Favors reactants

Watch Solution