What if the target equation has fractional coefficients?

Fractional coefficients are standard in thermochemistry, especially for formation reactions. Multiply intermediate equations by the necessary fractions to match these coefficients, and scale the enthalpy changes accordingly.

Can I use standard enthalpies of formation instead of reaction steps?

Yes. Using standard enthalpies of formation ($$\Delta H^\circ_f$$) is a specific application of Hess's law. You can calculate the total enthalpy change using the formula $$\Delta H^\circ_{rxn} = \Sigma n\Delta H^\circ_f( ext{products}) - \Sigma m\Delta H^\circ_f( ext{reactants})$$.

What happens if an intermediate species appears in more than two equations?

Focus on balancing the unique species first. The species appearing in multiple equations (commonly oxygen gas or water) will typically balance themselves out when the unique species are properly aligned with the target equation.

How to calculate enthalpy change with Hess's law

Hess's law states that the total enthalpy change for a chemical reaction is independent of the pathway taken, depending only on the initial and final states. It applies when determining the standard enthalpy of reaction ( $\Delta H^\circ_{rxn}$ ) for a target process using a set of intermediate reactions with known enthalpy changes.

The setup

Identify the target chemical equation for which the enthalpy change is unknown. Gather the provided intermediate chemical equations along with their corresponding standard enthalpy changes ( $\Delta H^\circ$ ). Ensure all chemical species include their state of matter (s, l, g, aq).

The steps

Manipulate the intermediate equations (by reversing them or multiplying by a scalar) so that the reactants and products match the target equation. 2. If you reverse an equation, flip the sign of its $\Delta H^\circ$ . 3. If you multiply an equation by a scalar, multiply its $\Delta H^\circ$ by the same scalar. 4. Add the manipulated equations together, canceling identical species that appear on both the reactant and product sides. 5. Add the adjusted $\Delta H^\circ$ values to find the total $\Delta H^\circ_{rxn}$ for the target equation.

Checking the result

Verify that the summation of the manipulated intermediate equations perfectly matches the target equation. All intermediate species must cancel out entirely. Ensure that species canceled across equations are in the exact same phase; for example, $H_2O(l)$ cannot cancel $H_2O(g)$ .

Common errors

A frequent error is forgetting to change the sign of $\Delta H^\circ$ when reversing an equation. Another is failing to multiply the $\Delta H^\circ$ value by the same scalar applied to the stoichiometric coefficients. Students also often incorrectly cancel species that are in different phases, leading to an incorrect net equation.

Worked example

Find $\Delta H^\circ_{rxn}$ for the reaction $2C(s) + H_2(g) \rightarrow C_2H_2(g)$ given the following intermediate equations: (1) $C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l)$ with $\Delta H_1 = -1299.5 ext{ kJ}$ ; (2) $C(s) + O_2(g) \rightarrow CO_2(g)$ with $\Delta H_2 = -393.5 ext{ kJ}$ ; (3) $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$ with $\Delta H_3 = -285.8 ext{ kJ}$ .

First, reverse equation (1) to put $C_2H_2(g)$ on the product side: $2CO_2(g) + H_2O(l) \rightarrow C_2H_2(g) + \frac{5}{2}O_2(g)$ with $\Delta H_1' = +1299.5 ext{ kJ}$ . Next, multiply equation (2) by 2 to supply the required $2C(s)$ for the reactants: $2C(s) + 2O_2(g) \rightarrow 2CO_2(g)$ with $\Delta H_2' = 2(-393.5) = -787.0 ext{ kJ}$ . Keep equation (3) as is to provide the $H_2(g)$ reactant: $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$ with $\Delta H_3' = -285.8 ext{ kJ}$ . Sum the manipulated equations: the $2CO_2(g)$ and $H_2O(l)$ produced in the intermediate steps cancel with the reactants in the reversed equation (1). The $2O_2(g)$ and $\frac{1}{2}O_2(g)$ combine to form $\frac{5}{2}O_2(g)$ , which cancels with the $\frac{5}{2}O_2(g)$ product. The net equation is $2C(s) + H_2(g) \rightarrow C_2H_2(g)$ . Finally, sum the adjusted enthalpies: $\Delta H^\circ_{rxn} = 1299.5 - 787.0 - 285.8 = +226.7 ext{ kJ}$ .

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References: Chemistry 2e, OpenStax, Chapter 5: Thermochemistry · Khan Academy, Thermodynamics: Hess's law and reaction enthalpy change