Question

The heat for the reaction ${N_2} + 3{H_2} \to 2N{H_3}$at ${27^ \circ }{\text{C}}$is $ - 91.94{\text{ kJ}}$. What will be its value at ${50^ \circ }{\text{C}}$? (The molar heat capacities at constant ${\text{P}}$ and ${27^ \circ }{\text{C}}$ for \[{N_2}\], \[{H_2}\] and \[N{H_3}\] are ${\text{28}}{\text{.45}}$, ${\text{28}}{\text{.32}}$ and ${\text{37}}{\text{.07 J }}{{\text{K}}^{ - 1}}$ respectively.
A) $ - 92.843{\text{ kJ}}$
B) $ - 47.7723{\text{ kJ}}$
C) $ - 132.5{\text{ kJ}}$
D) $ - 176.11{\text{ kJ}}$

Answer 1

Hint: To solve this we must know the expression for the molar heat capacity at constant pressure. First calculate the molar heat capacity at constant pressure for the reaction from the molar heat capacities at constant pressure for \[{N_2}\], \[{H_2}\] and \[N{H_3}\]. Then calculate the value of heat at ${50^ \circ }{\text{C}}$.

Formula Used:${C_P} = {\left( {\dfrac{{\Delta H}}{{\Delta T}}} \right)_P}$

Complete step-by-step solution:
We know the equation for molar heat capacity at constant pressure is as follows:
${C_P} = {\left( {\dfrac{{\Delta H}}{{\Delta T}}} \right)_P}$
Where, ${C_P}$ is the molar heat capacity at constant pressure.
$\Delta H$ is the change in heat.
$\Delta T$ is the change in temperature.

Rearrange the equation for the change in heat as follows:
$\Delta H = {C_P} \times \Delta T$
The change in heat is ${H_2} - {H_1}$ and the change in temperature is ${T_2} - {T_1}$. Thus,
${H_2} - {H_1} = {C_P}\left( {{T_2} - {T_1}} \right)$ …… (1)
Now we will calculate the molar heat capacity at constant pressure for the reaction as follows:
${C_P} = 2 \times {C_{P\left( {{\text{N}}{{\text{H}}_3}} \right)}} - \left[ {{C_{P\left( {{{\text{N}}_2}} \right)}} + \left( {3 \times {C_{P\left( {{{\text{H}}_2}} \right)}}} \right)} \right]$

We are given that the molar heat capacities at constant ${\text{P}}$ and ${27^ \circ }{\text{C}}$ for \[{N_2}\], \[{H_2}\] and \[N{H_3}\] are ${\text{28}}{\text{.45}}$, ${\text{28}}{\text{.32}}$ and ${\text{37}}{\text{.07 J }}{{\text{K}}^{ - 1}}$ respectively. Thus,
$\Rightarrow {C_P} = 2 \times \left( {{\text{37}}{\text{.07 J }}{{\text{K}}^{ - 1}}} \right) - \left[ {{\text{28}}{\text{.45 J }}{{\text{K}}^{ - 1}} + \left( {3 \times {\text{28}}{\text{.32 J }}{{\text{K}}^{ - 1}}} \right)} \right]$
$\Rightarrow {C_P} = \left( {74.14 - 113.41} \right){\text{ J }}{{\text{K}}^{ - 1}}$
$\Rightarrow {C_P} = - 39.27{\text{ J }}{{\text{K}}^{ - 1}}$
Thus, the molar heat capacity at constant pressure for the reaction is $ - 39.27{\text{ J }}{{\text{K}}^{ - 1}} = - 0.03927{\text{ kJ }}{{\text{K}}^{ - 1}}$.

Now, calculate the heat of the reaction at ${50^ \circ }{\text{C}}$ using equation (1) as follows:
We are given that the heat for the reaction at ${27^ \circ }{\text{C}}$ is $ - 91.94{\text{ kJ}}$. Thus,
$\Rightarrow {H_2} - \left( { - 91.94{\text{ kJ}}} \right) = - 0.03927{\text{ kJ }}{{\text{K}}^{ - 1}}\left( {50 - 27} \right)$
$\Rightarrow {H_2} + 91.94{\text{ kJ}} = - 0.03927{\text{ kJ }}{{\text{K}}^{ - 1}}\left( {23} \right)$
$\Rightarrow {H_2} = \left( { - 0.90321 - 91.94} \right){\text{ kJ}}$
$\Rightarrow {H_2} = - 92.843{\text{ kJ}}$
Thus, the heat of the reaction at ${50^ \circ }{\text{C}}$ is $ - 92.843{\text{ kJ}}$.

Thus, the correct answer is option (A) $ - 92.843{\text{ kJ}}$.

Note: We know that ${C_P}$ is the molar heat capacity at constant pressure. ${C_P}$ is the amount of heat absorbed or released by unit mass of a substance with change in temperature at constant pressure. The change in temperature causes a change in the enthalpy of the system. The molar heat capacity at constant pressure contributes to the work done as well as the change in internal energy. It is denoted by ${C_P}$.