OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीFor the formation reaction \( \text{H}_2(g) + \text{Br}_2(l) \rightarrow 2\text{HBr}(g) \), the reaction enthalpy is \( \Delta_r H^\ominus = -72.8\text{ kJ} \). The enthalpy change for the formation of 4.0 moles of \( \text{HBr}(g) \) from its elements in their standard reference states is:
Correct answer · Option 4
-145.6 kJ
Explanation
Molar enthalpy of formation \( \Delta_f H^\ominus(\text{HBr}) = -72.8 / 2 = -36.4\text{ kJ mol}^{-1} \). For 4.0 moles: \( \Delta H = 4.0 \times (-36.4) = -145.6\text{ kJ} \).
