OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीCalculate the standard enthalpy change for the reaction: \( \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \) using the bond enthalpy data: \( \text{B.E.}(\text{N}\equiv\text{N}) = 946\text{ kJ mol}^{-1} \), \( \text{B.E.}(\text{H}-\text{H}) = 435.8\text{ kJ mol}^{-1} \), and \( \text{B.E.}(\text{N}-\text{H}) = 389\text{ kJ mol}^{-1} \).
Correct answer · Option 2
\( -80.6\text{ kJ mol}^{-1} \)
Explanation
\( \Delta_r H^\ominus = [\text{B.E.}(\text{N}\equiv\text{N}) + 3\text{B.E.}(\text{H}-\text{H})] - [6\text{B.E.}(\text{N}-\text{H})] = [946 + 3(435.8)] - [6(389)] = [946 + 1307.4] - 2334 = 2253.4 - 2334 = -80.6\text{ kJ mol}^{-1} \).
