ΔH is path independent. Add/subtract known reactions to get target ΔH.

Hess's Law — Calculate ΔH of Reaction Using Given Data

Question

Given the following reactions:

\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \quad \Delta H_1 = -393.5 \text{ kJ/mol}

\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O(l)} \quad \Delta H_2 = -285.8 \text{ kJ/mol}

\text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)} \quad \Delta H_3 = -1366.8 \text{ kJ/mol}

Calculate the enthalpy of formation of ethanol, $\text{C}_2\text{H}_5\text{OH(l)}$ .

Solution — Step by Step

The enthalpy of formation means forming 1 mole of the compound from its elements in their standard states. So we need:

2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)} \quad \Delta H_f = ?

This is our target. Every step we take must result in this equation appearing.

We need 2C on the left — multiply Reaction 1 by 2. We need 3H₂O, so multiply Reaction 2 by 3. The ethanol combustion (Reaction 3) has ethanol on the left, but we need it on the right — so we reverse it.

\times 2: \quad 2\text{C(s)} + 2\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} \quad \Delta H = 2 \times (-393.5) = -787.0 \text{ kJ}

\times 3: \quad 3\text{H}_2\text{(g)} + \frac{3}{2}\text{O}_2\text{(g)} \rightarrow 3\text{H}_2\text{O(l)} \quad \Delta H = 3 \times (-285.8) = -857.4 \text{ kJ}

\text{Reversed}: \quad 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \quad \Delta H = +1366.8 \text{ kJ}

When we add these, species appearing on both sides cancel:

$2\text{CO}_2$ cancels (right of Rxn 1 vs left of reversed Rxn 3)
$3\text{H}_2\text{O}$ cancels (right of Rxn 2 vs left of reversed Rxn 3)
Oxygen: $2\text{O}_2 + \frac{3}{2}\text{O}_2 - 3\text{O}_2 = \frac{1}{2}\text{O}_2$ remains on the left ✓

What remains is exactly our target equation.

\Delta H_f = (-787.0) + (-857.4) + (+1366.8)

\Delta H_f = -1644.4 + 1366.8

\boxed{\Delta H_f = -277.6 \text{ kJ/mol}}

Why This Works

Hess’s Law is a direct consequence of enthalpy being a state function — it doesn’t matter which path a reaction takes, only the initial and final states determine $\Delta H$ . Think of it like altitude: climbing a mountain via the north face or south face, you gain the same height regardless of the route.

This means we can construct any reaction we want by algebraically combining reactions whose $\Delta H$ values we already know. The universe doesn’t care that ethanol “actually” forms from C, H₂, and O₂ in several messy steps — the enthalpy change is fixed.

This is why Hess’s Law is so powerful in thermochemistry: most formation reactions are impossible to carry out directly in a lab. Burning ethanol is easy to measure; synthesising it from pure carbon and hydrogen gas is not.

Alternative Method — Formation Enthalpy Formula

If you’re given standard enthalpies of formation ( $\Delta H_f°$ ) directly for all species in a reaction, use:

\Delta H_{reaction} = \sum \Delta H_f°(\text{products}) - \sum \Delta H_f°(\text{reactants})

For the combustion of ethanol (Reaction 3), rearranging this gives us $\Delta H_f°(\text{ethanol})$ directly. This is the same calculation, just framed differently — you’re still applying Hess’s Law under the hood.

In JEE Main, $\Delta H_f°$ of elements in their standard state (C(s) graphite, H₂(g), O₂(g)) is always zero. Don’t include them in the sum — they contribute nothing.

Common Mistake

The most common error here: forgetting to flip the sign when reversing a reaction. Students reverse Reaction 3 to get ethanol on the product side, but keep $\Delta H_3 = -1366.8$ kJ. The correct value after reversal is $+1366.8$ kJ. Reversing a reaction always reverses the sign — exothermic becomes endothermic and vice versa. This single error gives $\Delta H_f = -277.6 - 2 \times 1366.8$ , which is wildly off and should trigger a sanity check.

A quick sanity check: enthalpies of formation for most organic liquids are negative and in the range of $-100$ to $-500$ kJ/mol. If your answer is $+2000$ kJ/mol, something went wrong in the sign algebra.

Question

Solution — Step by Step

Why This Works

Alternative Method — Formation Enthalpy Formula

Common Mistake

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