Question

If $\Delta {{\text{H}}_{\text{f}}}^{\text{o}}\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right){\text{ = - 46}}{\text{.17kJ}}$ and $\Delta {{\text{H}}^{\text{o}}}$of the reaction ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}\left( {\text{g}} \right) \to {\text{2N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right)$is -187.6Kj, then $\Delta{\text{ H}}_{\text{f}}^{\text{o}}\left( {{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}} \right)$is:
(A)-95.4kJ
(B)+95.4kJ
(C)-141.5kJ
(D)-233.7kJ

Answer 1

Standard enthalpy change denoted by $\Delta{H^o}$ is the change in heat energy (enthalpy) during the formation of one mole of a substance from its elements. Enthalpy of formation is calculated from standard enthalpy of reaction.

Complete answer:
The enthalpy change of reaction or the standard enthalpy change of formation is given by the formula,$\Delta {H^ \circ }_{reaction} = \sum {\Delta H_f^ \circ } \left( {{\text{products}}} \right) - \sum {\Delta H_f^ \circ } \left( {{\text{reactants}}} \right)$. Where,
$\Delta {H^ \circ }_{reaction}$is the change in enthalpy of the reaction,
$\sum {\Delta H_f^ \circ } \left( {{\text{products}}} \right)$ is the sum of change in enthalpies of the products,
$\sum {\Delta H_f^ \circ } \left( {{\text{reactants}}} \right)$ is the sum of change in enthalpies of the reactants.
From the given equation ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}\left( {\text{g}} \right) \to {\text{2N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right)$, the reactant is ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ and the product is ${\text{N}}{{\text{H}}_{\text{3}}}$. Also the value of change in enthalpy of reaction is given as -187.6kJ, as only one product is obtained the change in its formation (of ammonia) is given by -46.17kJ. To find the enthalpy change of the reactant species i.e. hydrazine, let us substitute the given values of enthalpies in the above given equation.
Upon reframing the above formula using the given chemical species and substituting the value we get,
$\Delta {H^ \circ }_{reaction} = \sum {\Delta H_f^ \circ } \left( {{\text{2N}}{{\text{H}}_{\text{3}}}} \right) - \sum {\Delta H_f^ \circ } \left( {{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}} \right) \\\ \- 187.6 = \left( { - 46.17 \times 2} \right) - \sum {\Delta H_f^ \circ } \left( {{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}} \right) \\\ \sum {\Delta H_f^ \circ } \left( {{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}} \right) = 187.6 - 92.34 \\\ \sum {\Delta H_f^ \circ } \left( {{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}} \right) = 95.4 \\\$
From the calculation we got the value +95.4 Kilo joule.
So the answer for the question, enthalpy of formation of hydrazine is +95.4 Kj.

Note:
In the given question the number of moles of ammonia formed as a product is 2moles. So while calculating the value of enthalpy change of formation of ammonia is multiplied by two in the equation in order to get the correct answer.