Question

What is the enthalpy change (in kJ mol\(^{-1}\)) for the following reaction? \[ \mathrm{CH_4(g)\rightarrow C(g)+4H(g)} \] Given \[ \Delta_fH^\circ(\mathrm{CH_4})=-74.8\ \text{kJ mol}^{-1}, \] \[ \mathrm{H_2(g)\rightarrow 2H(g)};\qquad \Delta_aH^\circ=435\ \text{kJ mol}^{-1}, \] \[ \mathrm{C(s)\rightarrow C(g)};\qquad \Delta_aH^\circ=716.7\ \text{kJ mol}^{-1}. \]

• A. \(396.67\)
• B. \(1586.7\)
• C. \(1661.5\)
• D. \(415.37\)

Hint:

Use Hess's law: \[ \boxed{ \Delta H=\sum(\text{bond breaking/atomization})-\sum(\text{bond formation}) } \] Reverse the formation reaction when decomposing a compound into its elements.

Answer 1

The Correct Option is C

Step 1: Apply Hess's law. The required reaction is \[ \mathrm{CH_4(g)\rightarrow C(g)+4H(g)}. \] Break it into steps: \[ \mathrm{CH_4(g)\rightarrow C(s)+2H_2(g)} \] \[ \Delta H=+74.8\ \text{kJ mol}^{-1} \] (reverse of the enthalpy of formation). \[ \mathrm{C(s)\rightarrow C(g)} \] \[ \Delta H=716.7\ \text{kJ mol}^{-1}. \] \[ 2\mathrm{H_2(g)\rightarrow4H(g)} \] \[ \Delta H=2\times435=870\ \text{kJ mol}^{-1}. \]

Step 2: Calculate the total enthalpy change. \[ \Delta H = 74.8+716.7+870 = 1661.5\ \text{kJ mol}^{-1}. \]

Step 3: Write the answer. Hence, \[ \boxed{1661.5\ \text{kJ mol}^{-1}}. \] Thus, \[ \boxed{(C)} \] is the correct answer.