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Calculate the bond enthalpy of the \mathrm{\mathrm{N}-\mathrm{H}} bond in ammonia (NH3) using the following data:

Bond dissociation enthalpy of \mathrm{\mathrm{N}-\mathrm{H} } in hydrazine (N2H4): \mathrm{337 \mathrm{~kJ} / \mathrm{mol}} Bond dissociation enthalpy of N-N in hydrazine \mathrm{(N2H4): 167 \mathrm{~kJ} / \mathrm{mol}} Bond dissociation enthalpy of \mathrm{\mathrm{H}-\mathrm{H}} in hydrogen \mathrm{(\mathrm{H} 2): 436 \mathrm{~kJ} / \mathrm{mol}}

Option: 1

\mathrm{106 \mathrm{~kJ} / \mathrm{mol}}


Option: 2

\mathrm{266 \mathrm{~kJ} / \mathrm{mol}}


Option: 3

\mathrm{337 \mathrm{~kJ} / \mathrm{mol}}


Option: 4

\mathrm{557 \mathrm{~kJ} / \mathrm{mol}}


Answers (1)

best_answer

1. Write the balanced reaction equation:

                  \mathrm{ \mathrm{N} 2 \mathrm{H} 4(g) \rightarrow \mathrm{NH} 3(g)+\mathrm{H} 2(g) }
2. Calculate the bonds broken in the reaction and their respective enthalpies:

                Bond broken: \mathrm{N-N (in \, \, \mathrm{N} 2 \mathrm{H} 4)}
                  Enthalpy: 167 \mathrm{~kJ} / \mathrm{mol}

                  Bond broken: \mathrm{H-H (in \, \, N2H4)}

                  Enthalpy: \mathrm{436 \mathrm{~kJ} / \mathrm{mol}}

3. Calculate the bond formed in the reaction and its enthalpy:
                  Bond formed: \mathrm{N-H} (in NH3)
                  Enthalpy: ?

4. Apply Hess's law:

                \mathrm{ \Delta H=\Sigma\left(\Delta H_{\text {bonds broken }}\right)-\Sigma\left(\Delta H_{\text {bonds formed }}\right) }

5. Calculate \mathrm{\Delta H(N-H) :}

            \mathrm{ \Delta H(N-H)=(167 \mathrm{~kJ} / \mathrm{mol}+436 \mathrm{~kJ} / \mathrm{mol})-\Delta H }

6. Calculate \mathrm{\Delta H(N-H):}

               \mathrm{ \Delta H(N-H)=603 \mathrm{~kJ} / \mathrm{mol}-\Delta H }
7. Rearrange the equation to solve for \mathrm{\Delta H :}

             \mathrm{ \Delta H=603 \mathrm{~kJ} / \mathrm{mol}-\Delta H(N-H) }
8. Substitute the value of the bond dissociation enthalpy of \mathrm{\mathrm{N}-\mathrm{H}} in hydrazine (N2H4):

             \mathrm{ \Delta H=603 \mathrm{~kJ} / \mathrm{mol}-337 \mathrm{~kJ} / \mathrm{mol} }
9. Calculate the bond enthalpy of the \mathrm{\mathrm{N}-\mathrm{H}} bond in ammonia:

                            \mathrm{ \Delta H=266 \mathrm{~kJ} / \mathrm{mol} }

Answer: B) \mathrm{266 \mathrm{~kJ} / \mathrm{mol}}

Posted by

SANGALDEEP SINGH

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