Question:medium
What is the enthalpy change (in kJ mol\(^{-1}\)) for the following reaction?
\[
\mathrm{CH_4(g)\rightarrow C(g)+4H(g)}
\]
Given
\[
\Delta_fH^\circ(\mathrm{CH_4})=-74.8\ \text{kJ mol}^{-1},
\]
\[
\mathrm{H_2(g)\rightarrow 2H(g)};\qquad
\Delta_aH^\circ=435\ \text{kJ mol}^{-1},
\]
\[
\mathrm{C(s)\rightarrow C(g)};\qquad
\Delta_aH^\circ=716.7\ \text{kJ mol}^{-1}.
\]
What is the enthalpy change (in kJ mol\(^{-1}\)) for the following reaction?
\[
\mathrm{CH_4(g)\rightarrow C(g)+4H(g)}
\]
Given
\[
\Delta_fH^\circ(\mathrm{CH_4})=-74.8\ \text{kJ mol}^{-1},
\]
\[
\mathrm{H_2(g)\rightarrow 2H(g)};\qquad
\Delta_aH^\circ=435\ \text{kJ mol}^{-1},
\]
\[
\mathrm{C(s)\rightarrow C(g)};\qquad
\Delta_aH^\circ=716.7\ \text{kJ mol}^{-1}.
\]
Show 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.
Updated On: Jul 18, 2026
- \(396.67\)
- \(1586.7\)
- \(1661.5\)
- \(415.37\)
Show Solution
The Correct Option is C
Solution and Explanation
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