Hess’s law allows measuring** the enthalpy change (****Δ**** H)** for a reaction

**without making calorimetric measurements**. To do this, we need the enthalpies of some other reactions.

For example,

Using the given data for reactions (1) and (2), calculate the Δ*H* for the reaction:

2NO_{2}(*g*) + Cl_{2}(*g*) → 2NOCl(*g*) + O_{2}(*g*), **Δ***H*** = ?**

** __________________________________________________**

1) 2NO(g) + Cl_{2}(g) → 2NOCl(g), Δ*H* = -76 kJ

2) 2NO(g) + O_{2}(g) → 2NO_{2}(g), Δ*H* = -114 kJ

To find the Δ*H* for the target reaction, we need to **modify **equations 1 and 2 such that we **add them together and get the target equation**.

Of course, by doing so, we need to **follow rules** for manipulating thermochemical equations:

- If the equation is
**multiplied by any factor**, the**Δ**by the same factor.*H*must be multiplied - If the equation is
**reversed**, the**sign of Δ**.*H*must be changed

So, to get the target equation, first look up which of its components appear in the equations with known Δ*H *(equations 1 and 2).

NO_{2} is the first **reactant** in the target equation, and it appears in equation (2) as a **product**. Therefore, **reverse** the second reaction **to make NO _{2} a reactant** like in the target reaction. Let’s label the new equation as (2a). Make sure to change the sign for Δ

*H*:

2a) 2NO_{2}(g) → 2NO(g) + O_{2}(g), Δ*H* = +114 kJ

Pay attention to the coefficient as well; in both the target and (2a) equations, we have **2** NO_{2 }and this is what you want. The component in the reference equation most often needs to have the same coefficient as in the target equation.

The Cl_{2, }on the other hand, is a reactant in the target equation and so is it in equation (1), therefore, we keep **equation (1) intact**.

At this point, we can add equations (1) and (2a) to see if we get the target equation. Remember to add the enthalpies with correct signs as well:

1) 2NO(g) + Cl_{2}(g) → 2NOCl(g), Δ*H* = -76 kJ

+

2a) 2NO_{2}(g) → 2NO(g) + O_{2}(g), Δ*H* = +114 kJ

__________________________________________

2NO(g) + Cl_{2}(g) + 2NO_{2}(g) → 2NOCl(g) + 2NO(g) + O_{2}(g)

Δ*H* = -76 kJ + 114 kJ = +38 kJ

Cancel any molecules with the same coefficient that appear on both sides of the equation. In this case, it the NO:

2NO(g) + Cl_{2}(g) + 2NO_{2}(g) → 2NOCl(g) + 2NO(g) + O_{2}(g)

The final equation is:

2NO_{2}(g) + Cl_{2}(g) → 2NOCl(g) + O_{2}(g) Δ*H* = +38 kJ

And this matches the target equation, therefore, the Δ*H *was determined correctly.

**How does Hess’s Law Work?**

The secret souse of Hess’s law is that **enthalpy is a state function**, and therefore, Δ*H*_{rxn} is the same whether the reaction takes place in one step or in a series of steps.

Remember, a **state function** is a variable that only **depends on the initial and final states** and not how it was achieved. For example, altitude is a state function and wouldn’t matter how two hikers meet at a 500 ft altitude. One may be on his way to the top, and the other going back having behind a longer walk distance:

Now, by the same analogy, suppose we need to determine the enthalpy of the reaction 2A + B → 2D, given the enthalpies of the other two reactions:

2A + B → 2D **Δ***H*** = ?_____________________**

2A + B → C Δ*H* = +50 kJ

C → 2D Δ*H* = -80 kJ

On an energy diagram, the objective is to go from “2A + B” to “2D”, but we cannot directly measure the enthalpy change of this process. However, we know that if “2A + B” goes to C, it is +50 kJ, and “C” going to “2D” is -80 kJ.

And because enthalpy is a state function, it does not matter how “2A + B” goes to “2D”, so we add the enthalpy changes for these two routes/reactions to find the enthalpy change for going “2A + B” to “2D” directly:

So, to summarize, when solving a problem on the Hess’s law, remember your objective is to **manipulate the reference equations** such that when you **add them up**, the **target equation is obtained**. Remember, to multiply all the coefficients and the Δ*H* when needed, and to change the sign of ΔH when the equation is reversed.

**Check Also**

- Energy Related to Heat and Work
- Endothermic and Exothermic Processes
- Heat Capacity and Specific Heat
- Heat Capacity Practice Problems
- What is Enthalpy
- Constant-Pressure Calorimetry
- Bomb calorimeter – Constant Volume Calorimetry
- Stoichiometry and Enthalpy of Chemical Reactions
- Hess’s Law Practice Problems
- Standard Enthalpies of Formation
- Enthalpy of Reaction from Enthalpies of Formation
**Thermochemistry Practice Problems**

#### Practice

Calculate the enthalpy for the oxidation of CO to CO_{2} using the enthalpy of reaction for the combustion of C to CO (Δ*H* = -221.0 kJ) and the enthalpy for the combustion of C to CO_{2} (Δ*H* = -393.5 kJ).

2CO(g) + O_{2}(*g*) → 2CO_{2}(g) **Δ H = ?**

C(s) + O_{2}(*g*) → CO_{2}(g) Δ*H* = -393.5 kJ

2C(s) + O_{2}(*g*) → 2CO(g) Δ*H* = -221.0 kJ

-566 kJ

The target reaction contains 2 moles of CO_{2} as a product, so we need to multiply equation 1 by two and do the same for the value of Δ*H:*

1a) 2C(s) + 2O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -787 kJ

The second equation needs to be reversed since, in the target reaction, CO is a reactant while here it is a product:

2a) 2CO(g) → 2C(s) + O_{2}(*g*), Δ*H* = 221.0 kJ

And now we can add the two equations:

1a) 2C(s) + 2O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -787 kJ

+

2a) 2CO(g) → 2C(s) + O_{2}(*g*), Δ*H* = 221.0 kJ

2C(s) + 2O_{2}(*g*) + 2CO(g) → 2CO_{2}(g) + 2C(s) + O_{2}(*g*)

2CO(g) + O_{2}(*g*) → 2CO_{2}(g)

Δ*H* = -787 kJ + 221.0 kJ = -566 kJ

Calculate the enthalpy for the combustion reaction of sulfur to sulfur trioxide using the enthalpies of the two reactions shown below:

2S(*s*) + 3O_{2}(*g*) → 2SO_{3}(*g*) ** Δ H = ?**

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

2SO_{3}(*g*) → 2SO_{2}(*g*) + O_{2}(*g*) Δ*H* = 198 kJ

-792 kJ

The target reaction contains 2 moles of S as a reactant, so we need to multiply equation 1 by two and do the same for the value of Δ*H:*

1a) 2S(*s*) + 2O_{2}(*g*) → 2SO_{2}(*g*) Δ*H* = -594 kJ

The second equation needs to be reversed since in the target reaction, SO_{3} is a product while here it is a reactant:

2a) 2SO_{2}(*g*) + O_{2}(*g*) → 2SO_{3}(*g*) Δ*H* = -198 kJ

And now we can add the two equations:

1a) 2S(*s*) + 2O_{2}(*g*) → 2SO_{2}(*g*) Δ*H* = -594 kJ

+

2a) 2SO_{2}(*g*) + O_{2}(*g*) → 2SO_{3}(*g*) Δ*H* = -198 kJ

2S(*s*) + 2O_{2}(*g*) + 2SO_{2}(*g*) + O_{2}(*g*) → 2SO_{2}(*g*) + 2SO_{3}(*g*)

2S(*s*) + 3O_{2}(*g*) → 2SO_{3}(*g*)

Δ*H* = -594 kJ + (-198 kJ) = -792 kJ

Using the Hess’s law, calculate Δ*H*^{o} for the combustion reaction of butene:

C_{4}H_{8}(*g*) + 6O_{2}(*g*) → 4CO_{2}(*g*) + 4H_{2}O(*l*), **Δ***H*^{o }**= ?**

Use the following reactions and given Δ*H*’s:

1) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

2) C_{4}H_{8}(*g*) + H_{2}(*g*) → C_{4}H_{10}(*g*), Δ*H*^{o }= -126 kJ

3) 2C_{4}H_{10}(*g*) + 13O_{2}(*g*) → 8CO_{2}(*g*) + 10H_{2}O(*l*), Δ*H*^{o }= -5754 kJ

Δ*H*^{o }= -2718 kJ

Because C_{4}H_{8 }appears as a reactant in the target reaction, we are not going to reverse equation 2. However, to get rid of C_{4}H_{10, }we need to multiply equation 2 by two since there are two moles of C_{4}H_{10 }in equation 3. Let’s label the new equation as 2a:

2a) 2C_{4}H_{8}(*g*) + 2H_{2}(*g*) → 2C_{4}H_{10}(*g*), Δ*H*^{o }= -252 kJ

** **We cannot yet add equations 1, 2a, and 3 because there is going to be H2 only in the left side of the equation, but the target equation does not contain H2. This means we need to also reverse equation 1. Notice that reversing equation 2 for the same reason puts C_{4}H_{8 }as product and that is not what we have in the target reaction.

So, let’s reverse equation 1, changing the sign of the enthalpy, and label it as 1a:

1a) 2H_{2}O(*g*) → 2H_{2}(*g*) + O_{2}(*g*), Δ*H*^{o }= 571 kJ

Now, we can add equations 1a, 2a, and 3:

1a) 2H_{2}O(*g*) → 2H_{2}(*g*) + O_{2}(*g*), Δ*H*^{o }= 571 kJ

+

2a) 2C_{4}H_{8}(*g*) + 2H_{2}(*g*) → 2C_{4}H_{10}(*g*), Δ*H*^{o }= -252 kJ

+

3) 2C_{4}H_{10}(*g*) + 13O_{2}(*g*) → 8CO_{2}(*g*) + 10H_{2}O(*l*), Δ*H*^{o }= -5754 kJ

2H_{2}O(*g*) + 2C_{4}H_{8}(*g*) + 2H_{2}(*g*) + 2C_{4}H_{10}(*g*) + __13__O_{2}(*g*) → 2H_{2}(*g*) + O_{2}(*g*) + 2C_{4}H_{10}(*g*) + 8CO_{2}(*g*) + __10__H_{2}O(*l*)

2C_{4}H_{8}(*g*) + 12O_{2}(*g*) → 8CO_{2}(*g*) + 8H_{2}O(*l*)

Δ*H*^{o }= 571 kJ + (-252 kJ) + (-5754 kJ) = -5435 kJ

The last step is to simplify the equation by dividing it by two. Remember, the same goes for the enthalpy value:

2C_{4}H_{8}(*g*) + 12O_{2}(*g*) → 8CO_{2}(*g*) + 8H_{2}O(*l*), **Δ H^{o }= -2718 kJ**

Using the Hess’s law and the enthalpies of the three combustion reactions below, calculate the enthalpy of the reaction producing methanol (CH_{3}OH) from carbon monoxide and hydrogen gas.

CO(g) + 2H_{2}(*g*) → CH_{3}OH(g), **Δ H = ?**

1) 2CO(g) + O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -566 kJ

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

3) 2CH_{3}OH(g) + 3O_{2}(*g*) → 2CO_{2}(g) + 4H_{2}O(*g*), Δ*H*^{o }= −1430 kJ

-139 kJ

The first two equations seem to be in the correct directions since they contain CO and H_{2 }as reactants which is what we have in the target equation.

The **third equation**, however, **must be reversed** because the methanol here is a reactant while in the target equation, it is a product. So, let’s reverse it and label it as 3a:

3a) 2CO_{2}(g) + 4H_{2}O(*g*) → 2CH_{3}OH(g) + 3O_{2}(*g*), Δ*H*^{o }= +1430 kJ

Let’s add equations 1, 2, and 3a and see what we get:

1) 2CO(g) + O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -566 kJ

+

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

+

3a) 2CO_{2}(g) + 4H_{2}O(*g*) → 2CH_{3}OH(g) + 3O_{2}(*g*), Δ*H*^{o }= +1430 kJ

You may notice that there are four moles of water on the left side (equation 3a), but only two moles on the right side (equation 2). This means we need to **multiply equation two** **with its enthalpy value by two**. Let’s do that and label the new equation as 2a:

2a) 4H_{2}(*g*) + 2O_{2}(*g*) → 4H_{2}O(*g*), Δ*H*^{o }= -1141 kJ

And now, we can add equations 1, 2a, and 3a:

1) 2CO(g) + O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -566 kJ

+

2a) 4H_{2}(*g*) + 2O_{2}(*g*) → 4H_{2}O(*g*), Δ*H*^{o }= -1141 kJ

+

3a) 2CO_{2}(g) + 4H_{2}O(*g*) → 2CH_{3}OH(g) + 3O_{2}(*g*), Δ*H*^{o }= +1430 kJ

__________________________________________________

2CO) + O_{2} + 4H_{2} + 2O_{2} + 2CO_{2} + 4H_{2}O → 2CO_{2 }+ 4H_{2}O + 2CH_{3}OH + 3O_{2}

Cancel the same molecules on both side of the equation:

2CO + O_{2} + 4H_{2} + 2O_{2} + 2CO_{2} + 4H_{2}O → 2CO_{2} + 4H_{2}O + 2CH_{3}OH + 3O_{2}

Δ*H*^{o }= -566 kJ + (-1141 kJ) + 1430 kJ/mol = -277 kJ

So, we have:

2CO(g) + 4H_{2}(*g*) → 2CH_{3}OH(g), Δ*H*^{o }= -277 kJ

The last step is to **divide the equation by two** to match the target reaction:

CO(g) + 2H_{2}(*g*) → CH_{3}OH(g), Δ*H*^{o }= -139 kJ

Using the Hess’s law and the enthalpies of the given reactions, calculate the enthalpy of the combustion reaction of CH_{3}Cl:

CH_{3}Cl(*g*) + O_{2}(*g*) → CO(g) + HCl(*g*) + H_{2}O(*l*), **Δ H^{ }= ?**

1) CO(g) + 2H_{2}(*g*) → CH_{3}OH(g), Δ*H*^{o }= -139 kJ

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

3) CH_{3}OH(*g*) + HCl(*aq*) → CH_{3}Cl(*g*) + H_{2}O(*l*), Δ*H*^{o }= -28 kJ

-404 kJ

The first reaction needs to be reversed because the CO is a reactant, while in the target reaction it is a product. Let’s reverse it and label it as 1a:

1a) CH_{3}OH(g) → CO(g) + 2H_{2}(*g*), Δ*H*^{o }= +139 kJ

The second reaction is good as it is. However, the third reaction needs to be reversed as well in order to make CH_{3}Cl as a reactant as it is in the target equation:

3a) CH_{3}Cl(*g*) + H_{2}O(*g*) → CH_{3}OH(*g*) + HCl(*aq*), Δ*H*^{o }= +28 kJ

We can now add equations 1a, 2, and 3a:

1a) CH_{3}OH(g) → CO(g) + 2H_{2}(*g*), Δ*H*^{o }= +139 kJ

+

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

+

3a) CH_{3}Cl(*g*) + H_{2}O(*l*) → CH_{3}OH(*g*) + HCl(*aq*), Δ*H*^{o }= +28 kJ

____________________________________________

CH_{3}OH(g) + 2H_{2}(*g*) + O_{2}(*g*) + CH_{3}Cl(*g*) + 2H_{2}O(*l*) → CO(g) + 2H_{2}(*g*) + 2H_{2}O(*g*) + CH_{3}OH(*g*) + HCl(*aq*)

Cancel the same molecules on both side of the equation:

CH_{3}OH(g) + 2H_{2}(*g*) + O_{2}(*g*) + CH_{3}Cl(*g*) + H_{2}O(*l*) → CO(g) + 2H_{2}(*g*) + 2H_{2}O(*g*) + CH_{3}OH(*g*) + HCl(*aq*)

CH_{3}Cl(*g*) + O_{2}(*g*) → CO(g) + HCl(*g*) + H_{2}O(*g*)

Δ*H*^{o }= +139 kJ + (-571 kJ) + 28 kJ = -404 kJ

Using the Hess’s law and the enthalpies of the given reactions, calculate the enthalpy of the following oxidation reaction of ammonia:

4NH_{3}(g) + 5O_{2}(g) → 4NO(g) + 6H_{2}O(g), **Δ***H***= ?**

1) N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g), Δ*H* = -92 kJ

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H* = -484 kJ

3) N_{2}(g) + O_{2}(g) → 2NO(g), Δ*H* = 181 kJ

**-906 kJ**

The **first reaction needs to be reversed** because the NH_{3} is a product, while in the target reaction it is a reactant. Let’s reverse it and label it as 1a:

1a) 2NH_{3}(g) → N_{2}(g) + 3H_{2}(g), Δ*H* = +92 kJ

Let’s put the three equations together and see what the sum would look like:

1a) 2NH_{3}(g) → N_{2}(g) + 3H_{2}(g), Δ*H* = +92 kJ

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H* = -484 kJ

3) N_{2}(g) + O_{2}(g) → 2NO(g), Δ*H* = 181 kJ

One problem is that the hydrogens in equation 1a and 2 do not cancel each other since we have 3 moles in equation 1a and 2 moles in equation 2. Therefore, **let’s multiply equation 1a by two**, and **multiply equation 2 by three**:

1b) 4NH_{3}(g) → 2N_{2}(g) + 6H_{2}(g), Δ*H* = +184 kJ

2a) 6H_{2}(*g*) + 3O_{2}(*g*) → 6H_{2}O(*g*), Δ*H* = -1452 kJ

And now let’s put these three equations together:

1b) 4NH_{3}(g) → 2N_{2}(g) + 6H_{2}(g), Δ*H* = +184 kJ

2a) 6H_{2}(*g*) + 3O_{2}(*g*) → 6H_{2}O(*g*), Δ*H* = -1452 kJ

3) N_{2}(g) + O_{2}(g) → 2NO(g), Δ*H* = 181 kJ

The problem now is the number of moles of nitrogen in equations 1b and 3. We need to **multiply equation 3 by two:**

3a) 2N_{2}(g) + 2O_{2}(g) → 4NO(g), Δ*H* = 362 kJ

And now we can add equations 1b, 2a, and 3a:

1b) 4NH_{3}(g) → 2N_{2}(g) + 6H_{2}(g), Δ*H* = +184 kJ

+

2a) 6H_{2}(*g*) + 3O_{2}(*g*) → 6H_{2}O(*g*), Δ*H* = -1452 kJ

+

3a) 2N_{2}(g) + 2O_{2}(g) → 4NO(g), Δ*H* = 362 kJ

____________________________________________

4NH_{3}(g) + 6H_{2}(*g*) + 3O_{2}(*g*) + 2N_{2}(g) + 2O_{2}(g) → 2N_{2}(g) + 6H_{2}(g) + 6H_{2}O(*g*) + 2NO(g)

4NH_{3}(g) + 5O_{2}(*g*) → 6H_{2}O(*g*) + 2NO(g)

Δ*H* = +184 kJ + (-1452 kJ) + 362 kJ = **-906 kJ**

Using the Hess’s law and the enthalpies of the given reactions, calculate the enthalpy of the following oxidation reaction between CuO and HCl:

2CuO(s) + 4HCl(g) → 2CuCl(s) + Cl_{2}(g) + 2H_{2}O(g), **Δ***H***= ?**

1) CuO(s) + H_{2}(g) → Cu(s) + H_{2}O(g), Δ*H* = -85 kJ

2) 2Cu(s) + Cl_{2}(g) → 2CuCl(s), Δ*H* = -274 kJ

3) H_{2}(g) + Cl_{2}(g) → 2HCl(g), Δ*H* = -184 kJ

**-76 kJ**

We need to **multiply equation 1 by two** since there are two moles of CuO(s) in the target equation. Let’s label the new equation as 1a:

1a) 2CuO(s) + 2H_{2}(g) → 2Cu(s) + 2H_{2}O(g), Δ*H* = -170 kJ

Equation 2 seems to be fine, so let’s leave it as it is.

There are 2H_{2}(g) in equation 1a, and this means that we need to **reverse equation 3** and **multiply it by two** so that we can have 2H_{2}(g) as a product and cancel them with the ones in equation 1a:

3a) 4HCl(g) → 2H_{2}(g) + 2Cl_{2}(g), Δ*H* = 368 kJ

** **Let’s add equations 1a, 2, and 3a and see if they add up:

1a) 2CuO(s) + 2H_{2}(g) → 2Cu(s) + 2H_{2}O(g), Δ*H* = -170 kJ

2) 2Cu(s) + Cl_{2}(g) → 2CuCl(s), Δ*H* = -274 kJ

3a) 4HCl(g) → 2H_{2}(g) + 2Cl_{2}(g), Δ*H* = 368 kJ

2CuO(s) + 2H_{2}(g) + 2Cu(s) + Cl_{2}(g) + 4HCl(g) → 2Cu(s) + 2H_{2}O(g) + 2CuCl(s) + 2H_{2}(g) + 2Cl_{2}(g)

2CuO(s) + 4HCl(g) → 2CuCl(s) + 2H_{2}O(g) + Cl_{2}(g)

Δ*H* = -170 kJ + (-274 kJ) + 368 kJ = **-76 kJ**