Practice: For the reaction: N_{2} (g) + 2 O_{2} (g) ⇌ 2 NO_{2} (g), K_{c} = 8.3 x 10^{ -10} at 25^{°}C. What is the concentration of N_{2} gas at equilibrium when the concentration of NO_{2} is twice the concentration of O_{2} gas?

Subjects

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Chemical Equilibrium | 61 mins | 0 completed | Learn Summary |

ICE Chart | 52 mins | 0 completed | Learn |

Le Chatelier's Principle | 24 mins | 0 completed | Learn Summary |

The Reaction Quotient | 11 mins | 0 completed | Learn Summary |

Additional Guides |
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Equilibrium Expressions |

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

Concept #1: Understanding an ICE Chart

**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 #1: We have a solution where Ag(CN)_{2}** ^{–}**(g), CN

**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 #2: We place 2.5 mol of CO and 2.5 mol of CO_{3 }in a 10.0 L flask and let the system come to equilibrium. What will be the final concentration of CO_{2}? K = 0.47

Practice: For the reaction: N_{2} (g) + 2 O_{2} (g) ⇌ 2 NO_{2} (g), K_{c} = 8.3 x 10^{ -10} at 25^{°}C. What is the concentration of N_{2} gas at equilibrium when the concentration of NO_{2} is twice the concentration of O_{2} gas?

Example #3: When 0.600 atm of NO_{2} was allowed to come to equilibrium the total pressure was 0.875 atm. Calculate the K_{p }of the reaction.

**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 #4: An important reaction in the formation of acid rain listed below.

2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)

Initially, 0.023 M SO_{2} and 0.015 M O_{2} are mixed and allowed to react in an evacuated flask at 340 ** ^{o}**C. When an equilibrium is established the equilibrium amount of SO

Example #5: If K_{c} is 32.7 at 300^{o}C for the reaction below:

H_{2} (g) + Br_{2} (g) ⇌ 2 HBr (g)

What is the concentration of H_{2} at equilibrium if a 20.0 L flask contains 5.0 mol HBr initially?

**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.

Practice: At a given temperature the gas phase reaction: N_{2} (g) + O_{2} (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.

Example #6: Consider the following reaction:

COBr_{2 }(g) ⇌ CO (g) + Br_{2 }(g)

A reaction mixture initially contains 0.15 M COBr_{2}. Determine the equilibrium concentration of CO if K_{c} for the reaction at this temperature is 2.15 x 10** ^{-3}**.

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Concept #1: Understanding an ICE Chart

Example #1: We have a solution where Ag(CN)2 (g), CN (g), a...

Example #2: We place 2.5 mol of CO and 2.5 mol of CO3 in a 1...

Practice #1: For the reaction below,KC = 8.3 x 10-10 at 25oC...

Example #3: When 0.600 atm of NO2 was allowed to come to equ...

Example #4: An important reaction in the formation of acid r...

Example #5: If Kc is 32.7 at 300oC for the reaction below:
...

Practice #2: At a given temperature the gas phase reaction:
...

Example #6: The Quadratic Formula

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

Carbon monoxide gas reacts with hydrogen gas at elevated temperatures to form methanol according to this equation: CO (g) + 2 H2 (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?
0.50
3.57
1.7
5.4

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

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

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

The equilibrium constant for the reaction below at 25°C is 4.8 x 10 -6. Calculate the equilibrium concentration (mol/L) of Cl2 (g) if the initial concentration of ICI (g) is 1.33 mol/L. There is no I2 or Cl2 initially present. 2 ICI (g) ⇌ I2 (g) + Cl2 (g)a.) 2.9 x 10 -3b.) 5.8 x 10 -3c.) 3.2 x 10 -6d.) 6.4 x 10 -6e.) 343

What is the amount of "C" at equilibrium when 1 mole of A and 1 mole of B react according to the chemical reaction: 2 A + 2 B ⇌ C ?a. between 0 and 0.5 moleb. exactly 1 molec. between 0.5 and 2 molesd. exactly 2 molese. greater than two moles

If a 10.0 L vessel is charged with 0.482 mol N 2, 0.933 mol O2, calculate the concentration of each species at equilibrium.2 N2 + O2 ⇌ 2 N2O Kc = 2.0 x 10 -32

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

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 2 partial pressure was 2.8 atmospheres, what is Kp?
a) 0.28
b) 1.79
c) 2.87
d) 0.134
e) 3.54

Consider this reactionAB3(g) ⇌ A(g) + 3B(g) What is the equilibrium constant expression if the intial concentration of AB3 is 0.1 M and the equilibrium concentration of A is represented by x? Assume the intial concentrations of A and B are both zero.

Carbon monoxide gas reacts with hydrogen gas at elevated temperatures to form methanol according to this equation.CO (g) + 2 H2 (g) ⇋ CH3OH (g)When 0.40 mol of CO and 0.30 mol of H 2 are allowed to reach equilibrium in a 1.0 L container, 0.060 mol of CH3OH are formed. What is the value of Kc?a) 0.50b) 0.98c) 1.7d) 5.4

A mixture consisting of 0.250 M N2(g) and 0.500 M H2 (g) reaches equilibrium according to the equation N2(g) + 3 H2(g) → 2 NH3(g)At equilibrium, the concentration of ammonia is 0.150M. Calculate the concentration of H2(g) at equilibrium. 1. 0.350 M2. 0.150 M3. 0.425 M4. 0.275 M5. 0.0750 M

A flask is filled with 1.08 atm of oxygen gas and 0.88 atm of ammonia and the following reaction is allowed to come to equilibrium in the closed container.4 NH3 (g) + 5 O 2 (g) <==> 4 NO (g) + 6 H 2O (g)At equilibrium the pressure of O 2 is determined to be 0.38 atm. Calculate the pressure H 2O at equilibrium.A. 0.140 atmB. 0.840 atmC. 2.28 atmD. 0.456 atmE. 3.50 atm

Consider the following reaction at 425 KBr2 (g) + Cl2 (g) ⇌ 2 BrCl (g) Kc = 5.81If a 5.00 L reaction vessel in initially charged with 4.65 moles of Br 2 and 4.65 moles of Cl2, determine the concentration of Cl2 at equilibrium.A. 0.422 MB. 2.11 MC. 0.691 MD. 1.69 ME. 0.508 M

Consider the reaction shown below at 2200 °CN2 (g) + O2 (g) ↔ 2 NO (g) K c = 0.0505A student adds 0.375 mol of N2 and 0.375 mol of O2 to a 2.50 L flask and then allows the reaction to come to equilibrium. What are the equilibrium concentrations of all reactants and products?

A flask is filled with sulfuryl chloride, SO2Cl2 and has an initial concentration of 0.168 M. If the flask is allowed to come to equilibrium according to the reaction shown below, what is the concentration of SO2 in the flask at equilibrium?SO2Cl2 (g) <==> SO2 (g) + Cl2 (g) Kc = 2.32 x 10 -7A. 1.97 x 10-4 MB. 0.00851C. 3.90 x 10-8 MD. 4.39 x 10-4 ME. 1.62 x 10-4 M

Consider the following reaction:CH4 (g) + 2H2S (g) ↔ CS2 (g) + 4H2 (g)A reaction mixture initially contains 0.680 M CH 4 and 0.520 M H2S. When equilibrium is reached the concentration of H2 is determined to be 0.250 M. What equilibrium concentration of H 2S?a. 0.270 Mb. 0.583 Mc. 0.0200 Md. 0.395 Me. 0.458 M

Consider the following reaction:Fe3+(aq) + SCN-(aq) ⇌ FeSCN2+(aq)A solution is made containing an initial [Fe3+] of 1.1 x 10-3M an initial [SCN-] of 7.8 x 10-4M. At equilibrium, [FeSCN2+] = 1.7 x 10-4M.A) Kc = 2.0 x 102B) Kc = 3 x 10-2C) Kc = 3.3 x 10-3D) Kc = 3.0 x 102E) Kc = 5.0 x 10-3

A reaction mixture initially contains [N2O4] = 0.0250M. Find the equilibrium concentration of NO2.N2O4(g) ⇌ 2NO2(g) K c = 0.36A) 0.054 MB) 0.140 MC) 0.005 MD) 0.041ME) 0.027 M

Consider the following reaction:COCl2 (g) ⇌ CO (g) + Cl2 (g)A reaction mixture initially contains 1.6 M COCl 2. Determine the equilibrium concentration of CO if Kc for the reaction at this temperature is 8.33 x 10 -4. A) 4.2 x 10-4 MB) 1.5 x 10-3 MC) 3.7 x 10-2 MD) 2.1 x 10-2 ME) 1.3 x 10-3 M

A container initially contains [Br2] = 0.1 M and [I 2] = 0.2 M and undergoes the reaction:Br2 (g) + I 2 (g) ⇌ 2 BrI (g) Kc = 1x10–10 at 25 °C:Calculate the equilibrium concentration of BrI(g).A) 1x10−3B) 3.5x 10−6C) 1.4x10−6D) 7.7 x 10−7E) 0.012

A 5.0 L evacuated flask is charged with 2.0 moles of CO 2 and 1.5 moles of H 2 is allowed to come to equilibrium at 25 oC. If the equilibrium constant is 2.50, what is the equilibrium concentration of H2O? CO 2 (g) + H 2 (g) ⇌ CO (g) + H 2O (g)

The reaction NO2 (g) + NO (g) ⇌ N2O (g) + O2 (g) reached equilibrium at a certain high temperature. Originally the reaction vessel contained the following initial concentrations: 0.184 M N2O, 0.377 M O2 , 0.0560 M NO2 and 0.294 M NO. The concentration of NO2 , the only colored gas in the mixture, was monitored by following the intensity of the color. At equilibrium, the NO2 concentration had become 0.118 M. What is the value of Kc for the reaction at this temperature?

At a certain temperature, 0.0740 mol of PCl 5(g) were introduced into a one-liter container.PCl5(g) ⇋ PCl3(g) + Cl2(g)At equilibrium the concentration of PCl 3(g) was 0.0500 mol/L.a) What were the equilibrium concentrations of Cl 2(g) and PCl5(g)? b) What is the value of Kc at the temperature of the experiment?

Consider the reaction below:CaS (aq) + MgSO4 (aq) ↔ CaSO4 (s) + MgS (aq)If K is 20 and the initial concentrations of CaS, MgSO4 and MgS are 0.5 M, 0.6 M and 0 M respectively, what is the equilibrium concentration of MgSO4?1. 0.1 M2. 0.4 M3. 0.2 M4. 0 M

A 1.000 L vessel is filled with 2.000 moles of N 2, 1.000 mole of H2, and 2.000 moles of NH3. N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)When the reaction comes to equilibrium, it is observed that the concentration of H2 is 2.34 moles/L. What is the numerical value of the equilibrium constant Kc?

A 2.00 liter vessel is filled with NH 3 (g) to a pressure of 4.00 atm and the reactionN2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)comes to equilibrium at a fixed temperature. At equilibrium it is found that the pressure of H2 is 2.7 atm. What is the equilibrium pressure of NH3?

Practice: Consider the famous ammonia preparation3 H2(g) + N2(g) ⇌ 2 NH3(g)The equationK = [x]2 / [0.1 − 3x]3 [0.7 − x]is not a possible correct description of the equilibrium situation because

For the reaction: H 2 (g) + Br 2 (g) ⇌ 2 HBr (g), K 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

At a certain temperature, A2 and B2 react to give AB as follows:A2 (g) + B2 (g) ⇌ 2 AB (g) K c = 0.354If an 8.00 L flask initially contains 4.16 moles of A 2 and 4.16 moles of B2, find the equilibrium concentration of AB in the mixture.A. 0.142 MB. 0.175 MC. 0.119 MD. 1.91 ME. 0.238 M

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) 0.190B. 1.79C. 0.00193D. 2.46E. 0.596

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?

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 = 1.5x10−11A. 1.5x10−13 MB. 0.20 MC. 3.9x10−7 MD. 1.9x10−5 M

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation isS2(g) + C(s) → CS2(g) Kc = 9.40 at 900 KHow many grams of CS2(g) can be prepared by heating 11.1 moles of S2(g) with excess carbon in a 7.00 L reaction vessel held at 900 K until equilibrium is attained?

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?

The system described by the reactionCO(g) + Cl2(g) ⇌ COCl2(g)is at equilibrium at a given temperature when PCO =0.31 atm , PCl2 = 0.11 atm , and PCOCl2 = 0.59 atm. An additional pressure of Cl2(g) = 0.38 atm is added.Find the pressure of CO when the system returns to equilibrium.

For the exothermic reactionPCl3(g) + Cl2(g) ⇌ PCl5(g) Kp = 0.180 at a certain temperature.A flask is charged with 0.500 atm PCl3, 0.500 atm Cl2, and 0.300 atm PCl5 at this temperature.What are the equilibrium partial pressures of PCl 3, Cl2, and PCl5, respectively?Express your answers numerically in atmospheres with three digits after the decimal point, separated by commas.

Consider the following reaction: 2NO(g) + O 2(g) ⇌ 2NO2(g)A reaction mixture at 175 K initially contains 522 torr of NO and 421 torr of O2. At equilibrium, the total pressure in the reaction mixture is 748 torr. Calculate Kp at this temperature.

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.350 M, [B] = 1.40 M, and [C] = 0.700 M. The following reaction occurs and equilibrium is established: A + 2B ⇌ C At equilibrium, [A] = 0.170 M and [C] = 0.880 M. Calculate the value of the equilibrium constant, Kc. Express your answer numerically.

The reaction X2(g) ⇌ 2X(g) occurs in a closed reaction vessel at constant volume and temperature. Initially, the vessel contains only X2 at a pressure of 1.75 atm. After the reaction reaches equilibrium, the total pressure is 2.75 atm.What is the value of the equilibrium constant, Kp, for the reaction?

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?

At a certain temperature, the equilibrium constant K c for this reaction is 53.3.H2 + I 2 → 2 HIAt this temperature, 0.800 mol of H2 and 0.800 mol of I2 were placed in 1.00L container to react. What concentration of HI is present at equilibrium?

Consider the reaction between iodine gas and chlorine gas to form iodine monochloride: I2(g) + Cl2(g) ⇌ 2ICl(g); Kp = 81.9 (at 298 K)A reaction mixture at 298 K initially contains PI2 = 0.35 atm and PCl2 = 0.35 atm. What is the partial pressure of iodine monochloride when the reaction reaches equilibrium?

An equilibrium mixture contains 0.700 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g) How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

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?

For the reactionH2(g) + CO2(g) ⇌ H2O(g) + CO(g)at 700°C, 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 H2O is heated to 700°C in a 30.0-L container.

Consider the following reaction:SO2Cl2(g) ⇌ SO2(g) + Cl2(g)Kc = 2.99 × 10−7 at 227∘CIf a reaction mixture initially contains 0.177 M SO2Cl2, what is the equilibrium concentration of Cl 2 at 227∘C?

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.

A Student ran the following reaction in the laboratory at 502 K:PCl3(g) + Cl2(g) ⇌ PCl5(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 x 10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.

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?

Consider the following reaction. A (aq) ⇌ 2B (aq) K c = 4.70 x 10 -6 at 500 K If a 4.10 M sample of A is heated to 500 K, what is the concentration (M) of B at equilibrium?

The equilibrium constant Kc = 49 for the reaction: PCl3 (g) + Cl2 (g) ⇌ PCl5 (g) at 230 °C. If 0.70 mol of PCl3 is added to 0.70 mol of Cl2 in a 1.00-L reaction vessel at 230 °C, what is the concentration of PCl3 when equilibrium has been established? a. 0.59 M b. 0.30 M c. 0.11 M d. 0.83 M

At equilibrium, the concentrations in this system were found to be [N 2] = [O2] = 0.100 M and (NO) = 0.400 M. N2 (g) + O2 (g) ⇌ 2NO (g) If more NO is added, bringing its concentration to 0.700 M, what will the final concentration of NO be after equilibrium is re-established?

Ammonium Iodide dissociates reversibly to ammonia and hydrogen iodide:NH4I(s) ⇋ NH3(g) + HI(g)Kp = 0.215 at 400°CIf 150 g of ammonium iodide is placed into a 3.00 L vessel and heated to 400°C, calculate the partial pressure of ammonia when equilibrium is reached. Hint: Ammonium Iodide is a solid, is it used in the equilibrium calculation?

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