Enthalpies of formation of CCl₄( g), H₂ O(g), CO₂( g) and HCl(g) are -105,-242,-394 and -92 kJ mol⁻¹ respectively. The magnitude of enthalpy of the reaction given below is _________ kJ mol⁻¹. (nearest integer)
CCl₄( g)+2 H₂ O(g) → CO₂( g)+4 HCl(g)

Solution

Chemistry ma’am · 4 steps

Products minus reactants, and multiply each enthalpy of formation by its coefficient first.

1. The rule. The reaction enthalpy is the total enthalpy of formation of the products minus that of the reactants. Each value is multiplied by its stoichiometric coefficient.

   Δᵣ H = [Δ_f H(CO₂) + 4 Δ_f H(HCl)] - [Δ_f H(CCl₄) + 2 Δ_f H(H₂O)]

2. Products. -394 + 4 × (-92) = -394 - 368 = -762 kJ mol⁻¹.
3. Reactants. -105 + 2 × (-242) = -105 - 484 = -589 kJ mol⁻¹.
4. So. Δᵣ H = -762 - (-589) = -173 kJ mol⁻¹. The reaction is exothermic and the magnitude is 173.

The common slip: 139. Leaving out the coefficients 4 and 2 gives (-394 - 92) - (-105 - 242) = -139, so a magnitude of 139 instead of 173.

The idea to keep. For any reaction, Δᵣ H = Σ ν Δ_f H(products) - Σ ν Δ_f H(reactants). Never skip the coefficients ν.

Answer from the NTA final answer key. Solution by PrepWiser.