Question

A, B, C and D are some compounds. The enthalpy of formation of A(g), B(g), C(g) and D(g) is 9.7, -110, 81 and -393 kJ mol\(^{-1}\) respectively. What is \( \Delta H \) (in kJ mol\(^{-1}\)) for the given reaction? \[ A(g) + 3B(g) \rightarrow C(g) + 3D(g) \]

• A. \(-777.7\)
• B. \(+777.7\)
• C. \(-1418.3\)
• D. \(+1418.3\)

Hint:

To calculate enthalpy changes, apply Hess’s Law: - Use standard enthalpies of formation. - Subtract sum of reactant enthalpies from product enthalpies.

Answer 1

The Correct Option is A

Step 1: Using the enthalpy of formation formula: \[ \Delta H_{\text{reaction}} = \sum \Delta H_{\text{products}} - \sum \Delta H_{\text{reactants}} \] Substituting values: \[ \Delta H = [81 + 3(-393)] - [9.7 + 3(-110)] \] \[ = (81 - 1179) - (9.7 - 330) \] \[ = -1098 + 320.3 \] \[ = -777.7 \text{ kJ/mol} \]