Question

The heat of combustion of ${{\text{C}}_4}{{\text{H}}_{10}}$ is $- 2878{\text{kJ mo}}{{\text{l}}^{ - 1}}$ . If the heat of formation of ${\text{C}}{{\text{O}}_2}{\text{ and }}{{\text{H}}_2}{\text{O}}$ are $- 393.5{\text{kJmo}}{{\text{l}}^{ - 1}}{\text{ and }} - 285.8{\text{kJmo}}{{\text{l}}^{ - 1}}$ then the heat of formation of ${{\text{C}}_4}{{\text{H}}_{10}}$ is:
A.$- 125.0{\text{kJmo}}{{\text{l}}^{ - 1}}$
B.$126.75{\text{kJmo}}{{\text{l}}^{ - 1}}$
C.$- 402.5{\text{kJmo}}{{\text{l}}^{ - 1}}$
D.$402.5{\text{kJmo}}{{\text{l}}^{ - 1}}$

Answer 1

It can be solved using the methods which involve extraction of desired reaction or equation from the given reaction or equation by suitable addition, multiplication or subtraction.

Complete step by step answer:
Heat of combustion is defined as the amount of energy which is released when 1 mole of a substance is burnt completely with excess amount of oxygen. It is represented by $\Delta {\text{H}}_{\text{C}}^o$ . Heat of formation is the energy which is released or absorbed when 1 mole of a substance is formed from its elements in their most stable state. It is represented by $\Delta {\text{H}}_{\text{f}}^o$ . Heat of atomization is the energy which is absorbed by the 1 mole of a substance broken up to its isolated gaseous phase. It is represented by $\Delta {\text{H}}_{\text{a}}^o$ . Heat of fusion is defined as the energy which is absorbed by 1 mole of a solid to form liquid at its melting point. It is represented by $\Delta {\text{H}}_{\text{f}}^o$ .
As we have, heat of combustion of ${{\text{C}}_4}{{\text{H}}_{10}}$ is $- 2878{\text{kJ mo}}{{\text{l}}^{ - 1}}$ which means ${{\text{C}}_4}{{\text{H}}_{10}} + \dfrac{{13}}{2}{{\text{O}}_2} \to 4{\text{C}}{{\text{O}}_2} + 5{{\text{H}}_2}{\text{O}};\Delta {{\text{H}}_1} = - 2878{\text{kJmo}}{{\text{l}}^{ - 1}}$ . Let it to be reaction number 1.
The reverse of this equation can be given as: $4{\text{C}}{{\text{O}}_2} + 5{{\text{H}}_2}{\text{O}} \to {{\text{C}}_4}{{\text{H}}_{10}} + \dfrac{{13}}{2}{{\text{O}}_2};\Delta {{\text{H}}_2} = 2878{\text{kJmo}}{{\text{l}}^{ - 1}}$ . Let it to be reaction number 2.
As given, heat of formation of ${\text{C}}{{\text{O}}_2}{\text{ and }}{{\text{H}}_2}{\text{O}}$ are $- 393.5{\text{kJmo}}{{\text{l}}^{ - 1}}{\text{ and }} - 285.8{\text{kJmo}}{{\text{l}}^{ - 1}}$ , it can be given as: ${{\text{C}}_{\left( {\text{s}} \right)}} + {{\text{O}}_2} \to {\text{C}}{{\text{O}}_2};\Delta {{\text{H}}_3} = - 393.5{\text{kJmo}}{{\text{l}}^{ - 1}}$ and ${{\text{H}}_2} + \dfrac{1}{2}{{\text{O}}_2} \to {{\text{H}}_2}{\text{O}};\Delta {{\text{H}}_4} = - 285.8{\text{kJmo}}{{\text{l}}^{ - 1}}$ .
The heat of formation of ${{\text{C}}_4}{{\text{H}}_{10}}$ can be given as: $4{{\text{C}}_{\left( {\text{s}} \right)}} + 5{{\text{H}}_2} \to {{\text{C}}_4}{{\text{H}}_{10}};\Delta {\text{H}}$ .
This desired reaction can also be given as: $= {\text{equation}}2 + \left( {4 \times {\text{equation}}3} \right) + \left( {5 \times {\text{equation}}4} \right)$ .
Therefore heat of formation of ${{\text{C}}_4}{{\text{H}}_{10}}$ can be given as:
$\Delta H = \Delta {H_2} + \left( {4 \times \Delta {H_3}} \right) + \left( {5 \times \Delta {H_4}} \right)$ and putting values we get, $\Delta {\text{H}} = 2878 + \left( {4 \times - 393.5} \right) + \left( {5 \times - 285.8} \right)$ and on solving we get,
$\Delta {\text{H}} = - 125{\text{kJmo}}{{\text{l}}^{ - 1}}$ .

Thus, the correct option is A.

Note:
Heat of solution is the amount of heat change when 1 mole of a solute is dissolved in excess amount of solvent. Heat of neutralization is the energy which is released when one mole of water is formed by the neutralization of an acid or a base.