Question:
The atomization enthalpy of \(CH_4\) is \(1660\ \text{kJ mol}^{-1}\). The C-H bond enthalpy of each successive step in
\[
CH_4 \rightarrow CH_3 \rightarrow CH_2 \rightarrow CH \rightarrow C
\]
are \(+15\), \(+30\) and \(+45\ \text{kJ mol}^{-1}\) higher than the mean bond enthalpy of C-H bonds, respectively. The bond enthalpy of the last C-H unit is
The atomization enthalpy of \(CH_4\) is \(1660\ \text{kJ mol}^{-1}\). The C-H bond enthalpy of each successive step in
\[
CH_4 \rightarrow CH_3 \rightarrow CH_2 \rightarrow CH \rightarrow C
\]
are \(+15\), \(+30\) and \(+45\ \text{kJ mol}^{-1}\) higher than the mean bond enthalpy of C-H bonds, respectively. The bond enthalpy of the last C-H unit is
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The atomization enthalpy of a molecule is equal to the sum of all individual bond dissociation enthalpies present in the molecule.
Updated On: Jun 18, 2026
- \(400\ \text{kJ mol}^{-1}\)
- \(325\ \text{kJ mol}^{-1}\)
- \(475\ \text{kJ mol}^{-1}\)
- \(385\ \text{kJ mol}^{-1}\)
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The Correct Option is B
Solution and Explanation
Step 1: Calculate the mean C-H bond enthalpy.
The atomization enthalpy of methane is equal to the sum of all four C-H bond dissociation enthalpies. \[ \Delta H_{\text{atomization}}=1660\ \text{kJ mol}^{-1} \] Therefore, the mean C-H bond enthalpy is \[ \bar{D}=\frac{1660}{4} \] \[ =415\ \text{kJ mol}^{-1} \]
Step 2: Write the bond enthalpies of the first three successive steps.
Given that the first three successive bond dissociation enthalpies are respectively \(15\), \(30\) and \(45\ \text{kJ mol}^{-1}\) higher than the mean value. Hence, \[ D_1=415+15=430\ \text{kJ mol}^{-1} \] \[ D_2=415+30=445\ \text{kJ mol}^{-1} \] \[ D_3=415+45=460\ \text{kJ mol}^{-1} \]
Step 3: Calculate the last C-H bond enthalpy.
Let the last bond enthalpy be \(D_4\). Then, \[ D_1+D_2+D_3+D_4=1660 \] Substituting the values, \[ 430+445+460+D_4=1660 \] \[ 1335+D_4=1660 \] \[ D_4=1660-1335 \] \[ D_4=325\ \text{kJ mol}^{-1} \]
Step 4: Final conclusion.
Therefore, the bond enthalpy of the last C-H bond is \[ \boxed{325\ \text{kJ mol}^{-1}} \] Hence, the correct option is (2).
The atomization enthalpy of methane is equal to the sum of all four C-H bond dissociation enthalpies. \[ \Delta H_{\text{atomization}}=1660\ \text{kJ mol}^{-1} \] Therefore, the mean C-H bond enthalpy is \[ \bar{D}=\frac{1660}{4} \] \[ =415\ \text{kJ mol}^{-1} \]
Step 2: Write the bond enthalpies of the first three successive steps.
Given that the first three successive bond dissociation enthalpies are respectively \(15\), \(30\) and \(45\ \text{kJ mol}^{-1}\) higher than the mean value. Hence, \[ D_1=415+15=430\ \text{kJ mol}^{-1} \] \[ D_2=415+30=445\ \text{kJ mol}^{-1} \] \[ D_3=415+45=460\ \text{kJ mol}^{-1} \]
Step 3: Calculate the last C-H bond enthalpy.
Let the last bond enthalpy be \(D_4\). Then, \[ D_1+D_2+D_3+D_4=1660 \] Substituting the values, \[ 430+445+460+D_4=1660 \] \[ 1335+D_4=1660 \] \[ D_4=1660-1335 \] \[ D_4=325\ \text{kJ mol}^{-1} \]
Step 4: Final conclusion.
Therefore, the bond enthalpy of the last C-H bond is \[ \boxed{325\ \text{kJ mol}^{-1}} \] Hence, the correct option is (2).
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