Let B.E of X1, Y2 and XY are x kJ mol⁻¹, 0.5 x kJ mol⁻¹ and x kJ mol⁻¹ respectively. 1/2 X1 + 1/2 Y2 → XY; ΔH = −200 kJ mol⁻¹. Calculate ΔH.

Question

Accepted Answer

(c) ΔH = −200 + Σ (B.E)reactants − Σ (B.E)product. The enthalpy change (ΔH) is calculated as the sum of bond energies of reactants minus the sum of bond energies of products. Since ΔH is given as −200 kJ mol⁻¹, the correct formula is ΔH = −200 + Σ (B.E)reactants − Σ (B.E)product.

Let B.E of X1, Y2 and XY are x kJ mol⁻¹, 0.5 x kJ mol⁻¹ and x kJ mol⁻¹ respectively. 1/2 X1 + 1/2 Y2 → XY; ΔH = −200 kJ mol⁻¹. Calculate ΔH.

Solution

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