Hess's law — how to calculate ΔH using multiple reactions step by step

Question

Given the following thermochemical equations:

$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$ , $\Delta H_1 = -393.5 \text{ kJ/mol}$
$\text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$ , $\Delta H_2 = -283.0 \text{ kJ/mol}$

Find $\Delta H$ for the reaction: $\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}$

(CBSE Class 11 / JEE Main pattern)

Solution — Step by Step

Our target: $\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}$

We need to manipulate the given reactions to arrive at this target. Hess’s law says enthalpy change depends only on initial and final states — not the path taken.

Look at the target: C(s) is on the left (same as Reaction 1), CO(g) is on the right (but in Reaction 2, CO is on the left). So we need Reaction 1 as-is and Reaction 2 reversed.

flowchart TD
    A["Target: C + ½O₂ → CO"] --> B{"Which given reactions\nhave these species?"}
    B --> C["Rxn 1: C + O₂ → CO₂\n(keep as-is, has C on left)"]
    B --> D["Rxn 2: CO + ½O₂ → CO₂\n(reverse it, need CO on right)"]
    C --> E["Add manipulated reactions"]
    D --> E
    E --> F["Cancel common species\n(CO₂ and ½O₂ cancel)"]
    F --> G["Get target + ΔH"]

Reversed Reaction 2: $\text{CO}_2\text{(g)} \rightarrow \text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)}$ , $\Delta H_2' = +283.0 \text{ kJ/mol}$

When you reverse a reaction, the sign of $\Delta H$ flips. This is the most common step where students make errors.

Reaction 1: $\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$

Reversed Reaction 2: $\text{CO}_2\text{(g)} \rightarrow \text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)}$

Adding: $\text{CO}_2$ cancels on both sides, $\text{O}_2$ partially cancels ( $1 - \frac{1}{2} = \frac{1}{2}$ ).

\Delta H = \Delta H_1 + \Delta H_2' = -393.5 + 283.0 = \mathbf{-110.5 \text{ kJ/mol}}

Why This Works

Hess’s law is really just conservation of energy applied to chemical reactions. Enthalpy is a state function — it depends only on the initial and final states, not on how many intermediate steps you took. Whether carbon burns directly to CO or goes through CO₂ first, the total energy change for the same overall transformation is identical.

This means we can treat thermochemical equations like algebraic equations: reverse them, multiply them by constants, and add them up. The $\Delta H$ values follow the same operations.

Alternative Method — Using Formation Enthalpies

If you have standard enthalpies of formation ( $\Delta_f H^\circ$ ) for all species, use:

\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})

For this problem, $\Delta_f H^\circ[\text{C(s)}] = 0$ (element in standard state), $\Delta_f H^\circ[\text{O}_2] = 0$ , and $\Delta_f H^\circ[\text{CO}] = -110.5$ kJ/mol — which matches our answer.

In JEE Main, Hess’s law problems often give 3 reactions and ask you to find $\Delta H$ for a 4th. The algorithm is always the same: identify which species you need on which side, reverse/multiply as needed, then add. Practice 10 problems and you will crack any variant in under 3 minutes.

Common Mistake

The number one error: forgetting to flip the sign of $\Delta H$ when reversing a reaction. Students reverse the equation correctly but keep the original $\Delta H$ . If $\text{A} \rightarrow \text{B}$ has $\Delta H = -283$ kJ, then $\text{B} \rightarrow \text{A}$ has $\Delta H = +283$ kJ. Always. No exceptions.

A second common slip: multiplying a reaction by 2 but forgetting to multiply $\Delta H$ by 2 as well. If you double the coefficients, you double the enthalpy change.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Formation Enthalpies

Common Mistake

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