A chemical reaction usually doesn’t happen in one step, but through a series of steps. To determine the overall energy or heat involved we use **Hess’s Law**.

**Concept:** Understanding Hess's Law

Welcome back, guys. In this new video, we're going to take a look at heat summation.

We're going to say many reactions usually don't happen in one step. We're going to say a lot of reactions happen in multiple steps in order to get to our final products. We're going to say under Hess's law the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps.

This might be a little complicated and kind of abstract, so we'll do an example and see what I mean by this.

To determine the overall energy or heat of a reaction look to isolate each compound given within that chemical reaction.

**Concept:** Using Hess's Law

Now, Hess's law basically follows the same pattern that we're going to see in this example. In this example, it says for the following example calculate the unknown delta H value from the given delta H values of the other equations.

Basically, what we have to realize is, we need to find the delta H of this reaction here and the way we're able to do that is by manipulating these two reactions here in order for them to come out to this equation above. By manipulating them, we'll be able to find the delta H value we're missing. It seems kind of complicated, but it really is easy. All we have to do is go compound by compound and see where we find them. If things don't match up the way we see them, we have to just manipulate the equation so that they do.

Let's take a look here. We're going to take a look first at sulfur solid. We look on the two bottom equations and say where do I see sulfur solid. We're going to see sulfur solid right here and only here. We're going to say here it does not match what we have up here. Up here we only have one mole of sulfur solid. I need to find a way of converting this also into one mole of sulfur solid.

One way we can do that is by multiplying by two. Not just a sulfur solid, but the entire equation. When we multiply everyone by two, all those halves become one. One mole of sulfur solid plus one mole of oxygen gas gives me one mole of SO2 gas.

Here we're going to say, if we multiply the equation times two, then we're also going to have to multiply delta H times two. Then that becomes negative 593.6 kilojoules. Remember, how do we do that? We just multiplied it by two. Because I'm manipulating that equation, it's no longer used. It's gone because it's changing to this new one.

Now, I have one mole of sulfur solid, just like I have up above, so we're good.

Next, we're going to look at oxygen gas, O2. Where do I see O2? I see O2 in two places. I see it here in the second equation and I see it here in my brand new first equation. When I find it in more than one place, that means I'm going to have to skip it because it's kind of hard to gauge what we should do next. If you find a compound in more than one place, skip it and move on to the next example.

Now we have SO3 as a product and there's only one mole of it. Here we have SO3 down here, but the problem – there's two problems with it. One, it's a reactant, but we need it to be a product and two, there's two moles of it.

What we're going to do first is we're going to reverse the reaction so that it's also a product just like up above. We're going to reverse the entire equation. So my reactants now become products and my products become reactants. Everything gets reversed. Just realize this, any time I reverse a reaction, that reverses the sign for delta H. Delta H was positive, now it becomes negative.

Now we have 2SO3 as a product, but we only need one mole of SO3 because that's what we have up above here. We only have one mole of SO3. So what I need to do lastly is I'm going to divide by two because we need to get rid of that two. So this becomes SO2 gas plus half O2 gas gives me SO3 gas.

Of course, if we divide the equation by a number, then we also have to divide delta H by that same number. This is going to get divided by two, so now my new delta H is negative 99.2 kilojoules. These two got manipulated, so they're both gone.

What I'm going to do next is I'm going to get rid of what are called intermediates. Intermediates are compounds that look alike, but one is a reactant in one equation and the other is a product in another equation. Our intermediate here is SO2. Here in this first equation, it's a product but in the second one, it's a reactant. Because one is a reactant and one is a product, they're going to cancel each other out. So now they're totally gone.

Now all we're going to do here is we're going to bring down everything that didn't get canceled out. The sulfur solid didn't get canceled out, so it comes down. This one mole of O2 combines with this half O2. They're both reactants, so they don't cancel out. They add up. So one 1 plus half gives me 1.5, which is the same thing as 3 over 2. Then this SO3 comes down.

If you notice, the equation we just isolated is the same equation up above. We're going to say that to find the delta H value of this equation, which we're going to say is delta H 3, all you do is you're just going to combine delta H 1 plus delta H 2. Just add them up. It would be negative – I'm going to take myself out of this image guys, so we can see the work. So it's going to be negative 593.6 kilojoules plus 99.2 kilojoules. That gives us negative 692.8 kilojoules.

Now just realize, all we have to do is we have to manipulate the equations with given delta H values to find the delta H of this reaction up here. That's the approach we need to take. So go compound by compound and see where you find them. Make sure you have to manipulate them by either multiplying, dividing, or reversing. Doing one of those three steps will help us isolate this equation. Once we do that, just add up your delta H values together.

Hess's law can be challenging, but always take that approach to help you better organize the information.

**Problem:** Calculate the ?Hrxn for CO(g) + NO (g) -----> CO2 (g) + N2 (g) ?H = ? Given the following set of reactions: CO2 (g) -----> CO (g) + 1/2 O2 (g) ?H1 = 283.0 kJ N2 (g) + O2 (g) ----> 2 NO (g) ?H2 = 180.6 kJ

**Problem:** Calculate the ?Hrxn for ClF (g) + F2 (g) -----> ClF3 (g) ?Hrxn = ? Given the following reactions: Cl2O (g) + F2O -----> 2 ClF (g) + O2 (g) ?Hrxn = - 167.4 kJ 4 ClF3 (g) + 4 O2 (g) -----> 2 Cl2O (g) + 6 F2O (g) ?Hrxn = 682.8 kJ 2 F2 (g) + O2 (g) -----> 2 F2O (g) ?Hrxn = -181.7 kJ

Given the following data:

3CO_{2}(g) + 4H_{2}O(l) ⇌ C_{3}H_{8}(g) + 5O_{2}(g)

∆H = 1, 110 kJ · mol^{−1}

H_{2}O(l) ⇌ H_{2}(g) + 1/2O_{2}(g)

∆H = 142.5 kJ · mol^{−1}

3C_{graphite} + 4H_{2}(g) ⇌ C_{3}H_{8}(g)

∆H = −52 kJ · mol^{−1}

calculate ∆H for the reaction

Cgraphite + O_{2}(g) ⇌ CO_{2}(g)

1. 1202 kJ · mol^{−1}

2. −592 kJ · mol^{−1}

3. 543 kJ · mol^{−1}

4. −197 kJ · mol^{−1}

5. −543 kJ · mol^{−1}

Watch Solution

S(s) + O_{2}(g) → SO_{2} (g) ∆H = ‐296.8 kJ

2SO_{2} + O_{2} → 2SO_{3} ∆H = ‐198.4 kJ

Write the equation for the formation of sulfur trioxide gas from sulfur solid and oxygen. What is the ∆H of that equation?

Watch Solution

Use the standard reaction enthalpies given below to determine ΔH°_{rxn} for the following reaction:

P_{4}(g) + 10Cl_{2}(g) → 4PCl_{5}(s) ΔH°_{rxn }= ?

Given

PCl_{5}(s) → PCl_{3}(g) + Cl_{2}(g) ΔH°_{rxn }= +157kJ

P_{4}(g) + 6Cl_{2}(g) → 4PCl_{3}(g) ΔH°_{rxn }= -1207kJ

A) -1835 kJ

B) - 1364 kJ

C) -1050 kJ

D) -1786kJ

E) -2100 kJ

Watch Solution

The values for the enthalpy of reaction are given for two reactions below

Mg(s) + Na _{2}CO_{3}(s) → 2 Na(s) + MgCO _{3}(s) ΔH _{rxn} = 317.1 kJ

Mg(s) + 2 NaF(s) → 2 Na(s) + MgF _{2}(s) ΔH _{rxn} = 35.6 kJ

Based on this information, what is the value for ΔH _{rxn} for the reaction

MgCO_{3}(s) + 2 NaF(s) → Na _{2}CO_{3}(s) + MgF_{2}(s)

a. + 352.7 kJ

b. + 281.5 kJ

c. + 245.9 kJ

d. - 281.5 kJ

e. - 352.7 kJ

Watch Solution

Given the chemical reactions:

2 NO_{2}(*g*) → N_{2}(*g*) + 2 O_{2}(*g*) ΔH = −68 kJ

N_{2}O_{4}(*g*) → N_{2}(*g*) + 2 O_{2}(*g*) ΔH = −10 kJ

Determine the ΔH for the following reaction:

N_{2}O_{4}(g) → 2 NO_{2}(*g*)

A. −78 kJ

B. −58 kJ

C. 48 kJ

D. 58 kJ

E. 78 kJ

Watch Solution

The United States gets 95% of its hydrogen gas by "steam reforming" methane, CH_{4}(g).

Another option is the “partial oxidation” of hydrocarbons to make carbon monoxide and hydrogen gas. For octane, that reaction would be

C_{8}H_{18}(l) + 4O_{2}(g) → 8CO(g) + 9H_{2}(g)

We know ΔH° for the combustion of octane

C_{8}H_{18}(l) + 25/2 O_{2}(g) → 8CO_{2}(g) + 9H_{2}O(l) ΔH° = -5500 kJ/mol

Assume the arbitrary values below for the combustion reactions of carbon monoxide and hydrogen:

CO(g) + 1/2 O_{2}(g) → CO_{2}(g) ΔH° = “-290 kJ/mol”

H2(g) + 1 /2O_{2}(g) → H_{2}O(l) ΔH° = “-285 kJ/mol”

A. Do you expect the magnitude of ΔH° for the partial oxidation of octane, to be greater or less than the magnitude of ΔH° for the combustion of octane? Explain.

**B. Now, determine ΔH° for the partial oxidation of octane with those assigned values. Choose from the choices below**

C. Do you think the hydrogen produced provides more or less energy than you would get from the combustion of octane? Explain.

Watch Solution

Calculate ΔH° (in kJ) for the reaction.

2S (s) + 3O_{2} (g) → 2SO_{3} (g) ∆H = -790 kJ

S (s) + O_{2} (g) → SO_{2} (g) ∆H = -297 kJ

the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide is ________ kJ.

a) -196

b) -543

c) 1087

d) 196

d) -1384

Watch Solution

Given the following reactions

N_{2} (g) + 2O_{2} (g) → 2NO_{2} (g) ΔH = 66.4 kJ

2NO (g) + O_{2} (g) → 2NO_{2} (g) ΔH = -114.2 kJ

the enthalpy of the reaction of the nitrogen to produce nitric oxide is ________ kJ.

a) -47.8

b) 47.8

c) 180.6

d) -180.6

e) 90.3

Watch Solution

Given the following reactions

Fe_{2}O_{3} (s) + 3CO (s) → 2Fe (s) + 3CO_{2} (g) ∆H = -28.0 kJ

3Fe (s) + 4CO_{2} (s) → 4CO (g) + Fe _{3}O_{4} (s) ∆H = +12.5 kJ

the enthalpy of the reaction of Fe _{2}O_{3} with CO

3Fe_{2}O_{3} (s) + CO (g) → CO_{2} (g) + 2Fe_{3}O_{4} (s)

is ________ kJ.

a) 40.5

b) +109

c) -15.5

d) -109

e) -59.0

Watch Solution

Researchers at Purdue University have exploited the reactivity of a reaction to develop a novel type of rocket fuel consisting of aluminum nanoparticles dispersed in H_{2}O(s). They nicknamed this propellant “ALICE” (get it? Al-ice) and noted that in addition to being environmentally friendly, ALICE could be produced on any extraterrestrial body that contains water. The recent discovery of water on Mars could draw new attention to this intriguing fuel. (International Journal of Aerospace Engineering, 2012)

One can speculate that the reaction that causes ALICE to act as a propellant is:

2Al (s) + 3H_{2}O(s) → Al_{2}O_{3}(s) + 3H_{2}(g)

where the hydrogen gas is subsequently burned.

A. Given the molar enthalpies for the following reactions,

2Al(s) + 3/2 O_{2} (g) → Al_{2}O_{3} (s) ΔH° = -1669.8 kJ/mol

H_{2} (g) + 1/2 O_{2} (g) → H_{2}O (l) ΔH° = -285.8 kJ/mol

H_{2}O (s) → H_{2}O (l) ΔH° = 6.01 kJ/mol

determine the ΔH° for the ALICE reaction

2Al (s) + 3H_{2}O (s) → Al_{2}O_{3} (s) + 3H_{2} (g)

Is the ΔH° you calculated consistent with Al-ice serving as a fuel? ________________

Give one reason to support your answer.

Watch Solution

Using Hess’s Law, determine the ΔH of the following reaction:

N_{2}(g) + 2O_{2}(g) → 2NO_{2}(g)

Given the following equations

N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g) ΔH = -115 kJ

2NH_{3}(g) + 4H_{2}O(l) → 2NO_{2}(g) + 7H_{2}(g) ΔH = -142.5 kJ

H_{2}O(l) → H_{2}(g) + 1/2O_{2}(g) ΔH = -43.7 kJ

Watch Solution

The heat of solution is LiCl is –37.1 kJ/mol, and the lattice energy of LiCl(s) is 828 kJ/mol. Calculate the total heat of hydration of 1 mol of gas phase Li^{+} ions and Cl ^{-} ions.

1) 791 kJ

2) 865 kJ

3) -865 kJ

4) -791 kJ

5) None of these

Watch Solution

Calculate the enthalpy of the following reaction:

Watch Solution

Determine the answer for the question below.

Watch Solution