Question:

The enthalpy of atomization of \({CH}_3{NH}_2(g)\) is 2313 kJ mol\(^{-1}\). If \(\Delta_{{C-H}}^{\ominus}\) and \(\Delta_{{N-H}}^{\ominus}\) are 414 and 389 kJ mol\(^{-1}\) respectively, then \(\Delta_{{C-N}}^{\ominus}\) (in kJ mol\(^{-1}\)) will be:

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Use bond enthalpy relations and atomization enthalpy to find unknown bond enthalpies.
Updated On: Jun 2, 2025
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The Correct Option is A

Solution and Explanation

Given: \[ \Delta_{{atomization}} = \Delta_{{C-H}} + \Delta_{{N-H}} + \Delta_{{C-N}} \] Values: \[ 2313 = 3 \times 414 + 2 \times 389 + \Delta_{{C-N}} \] Calculate: \[ 3 \times 414 = 1242, 2 \times 389 = 778 \] Sum: \[ 1242 + 778 = 2020 \] So, \[ \Delta_{{C-N}} = 2313 - 2020 = 293 \, {kJ mol}^{-1} \]
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