Hess's law — calculate enthalpy of formation from combustion data

medium CBSE JEE-MAIN • JEE Main 2022 • 4 min read
Tags Thermodynamics

Question

Using the following combustion data, calculate the standard enthalpy of formation of methane, CH4(g)\text{CH}_4(g):

ΔHc∘[C(graphite)]=−393.5 kJ/mol\Delta H^\circ_c[\text{C(graphite)}] = -393.5 \text{ kJ/mol} ΔHc∘[H2(g)]=−285.8 kJ/mol\Delta H^\circ_c[\text{H}_2(g)] = -285.8 \text{ kJ/mol} ΔHc∘[CH4(g)]=−890.3 kJ/mol\Delta H^\circ_c[\text{CH}_4(g)] = -890.3 \text{ kJ/mol}

Find ΔHf∘[CH4(g)]\Delta H^\circ_f[\text{CH}_4(g)].

Solution — Step by Step

The enthalpy of formation of CH4\text{CH}_4 means forming 1 mole of CH4\text{CH}_4 from its elements in standard states:

C(graphite)+2H2(g)→CH4(g)ΔHf∘= ?\text{C(graphite)} + 2\text{H}_2(g) \rightarrow \text{CH}_4(g) \quad \Delta H^\circ_f = \, ?

This is what we need to construct using Hess’s law.

Combustion always means reacting with O2\text{O}_2 to form CO2\text{CO}_2 and H2O\text{H}_2\text{O}:

(i)C(graphite)+O2(g)→CO2(g)ΔH=−393.5 kJ/mol\text{(i)}\quad \text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5 \text{ kJ/mol} (ii)H2(g)+12O2(g)→H2O(l)ΔH=−285.8 kJ/mol\text{(ii)}\quad \text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H = -285.8 \text{ kJ/mol} (iii)CH4(g)+2O2(g)→CO2(g)+2H2O(l)ΔH=−890.3 kJ/mol\text{(iii)}\quad \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -890.3 \text{ kJ/mol}

We need C\text{C} and H2\text{H}_2 on the left and CH4\text{CH}_4 on the right. Check:

  • Equation (i): C\text{C} is on the left — keep as is.
  • Equation (ii): H2\text{H}_2 is on the left — keep, but we need 2 mol of H2\text{H}_2, so multiply by 2.
  • Equation (iii): CH4\text{CH}_4 is on the left — we need it on the right, so reverse it.

Reversing a reaction flips the sign of ΔH\Delta H. This is the entire mechanism behind Hess’s law.

 (i)C+O2→CO2ΔH1=−393.5\text{(i)}\quad \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \Delta H_1 = -393.5 (ii)×22H2+O2→2H2OΔH2=2×(−285.8)=−571.6\text{(ii)} \times 2\quad 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \quad \Delta H_2 = 2 \times (-285.8) = -571.6 (iii reversed)CO2+2H2O→CH4+2O2ΔH3=+890.3\text{(iii reversed)}\quad \text{CO}_2 + 2\text{H}_2\text{O} \rightarrow \text{CH}_4 + 2\text{O}_2 \quad \Delta H_3 = +890.3

Add all three. The CO2\text{CO}_2, 2H2O2\text{H}_2\text{O}, and 2O22\text{O}_2 cancel from both sides.

 ΔHf∘=−393.5+(−571.6)+890.3\Delta H^\circ_f = -393.5 + (-571.6) + 890.3 ΔHf∘=−393.5−571.6+890.3=−74.8 kJ/mol\Delta H^\circ_f = -393.5 - 571.6 + 890.3 = \mathbf{-74.8 \text{ kJ/mol}}

The standard enthalpy of formation of CH4(g)\text{CH}_4(g) is −74.8-74.8 kJ/mol.

Why This Works

Hess’s law is really just the conservation of energy applied to chemistry. Enthalpy is a state function — it doesn’t matter which path a reaction takes, only the initial and final states matter. So we’re free to add, subtract, and scale equations as long as the algebra is clean.

The combustion data gives us a “bridge” through CO2\text{CO}_2 and H2O\text{H}_2\text{O}. We use those as intermediate species that cancel out, leaving only the formation reaction we actually want. Think of it as solving simultaneous equations — you’re eliminating variables you don’t need.

This exact approach — using combustion enthalpies to back-calculate formation enthalpies — is a standard JEE Main question type. The numbers change, but the strategy is always the same three moves: identify the target equation, write combustion equations, then reverse and scale to match.

Alternative Method

When combustion data is given, we can use the direct formula:

ΔHf∘[product]=∑ΔHc∘[reactants]−ΔHc∘[product]\Delta H^\circ_f[\text{product}] = \sum \Delta H^\circ_c[\text{reactants}] - \Delta H^\circ_c[\text{product}]

For CH4=C+2H2\text{CH}_4 = \text{C} + 2\text{H}_2:

ΔHf∘[CH4]=ΔHc∘[C]+2ΔHc∘[H2]−ΔHc∘[CH4]\Delta H^\circ_f[\text{CH}_4] = \Delta H^\circ_c[\text{C}] + 2\Delta H^\circ_c[\text{H}_2] - \Delta H^\circ_c[\text{CH}_4] =−393.5+2(−285.8)−(−890.3)=−393.5−571.6+890.3=−74.8 kJ/mol= -393.5 + 2(-285.8) - (-890.3) = -393.5 - 571.6 + 890.3 = -74.8 \text{ kJ/mol}

This shortcut works because the formula is derived from the same algebraic cancellation we did in Step 3–4. Once you’ve done the full method a few times, use this formula directly to save time in the exam.

Common Mistake

Forgetting to reverse the sign when reversing equation (iii).

The combustion of CH4\text{CH}_4 is exothermic (ΔH=−890.3\Delta H = -890.3 kJ/mol). When we reverse the equation so CH4\text{CH}_4 appears as a product, the reaction becomes endothermic: ΔH=+890.3\Delta H = +890.3 kJ/mol.

Students who keep the sign as −890.3-890.3 get ΔHf∘=−1855.4\Delta H^\circ_f = -1855.4 kJ/mol — a nonsensically large value that should immediately raise a red flag. If your answer has a magnitude above ~500 kJ/mol for a simple organic compound, recheck your signs.

A quick sanity check: formation enthalpies of simple hydrocarbons like CH4\text{CH}_4, C2H6\text{C}_2\text{H}_6, C2H4\text{C}_2\text{H}_4 are all in the range of −50-50 to +230+230 kJ/mol. If your answer is way outside this range, there’s a sign error somewhere in the manipulation.

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