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

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

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

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

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

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

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