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Question: The heat for the reaction \({N_2} + 3{H_2} \to 2N{H_3}\)at \({27^ \circ }{\text{C}}\)is \( - 91.94{\...

The heat for the reaction N2+3H2→2NH3{N_2} + 3{H_2} \to 2N{H_3}at 27∘C{27^ \circ }{\text{C}}is −91.94 kJ - 91.94{\text{ kJ}}. What will be its value at 50∘C{50^ \circ }{\text{C}}? (The molar heat capacities at constant P{\text{P}} and 27∘C{27^ \circ }{\text{C}} for N2{N_2}, H2{H_2} and NH3N{H_3} are 28.45{\text{28}}{\text{.45}}, 28.32{\text{28}}{\text{.32}} and 37.07 J K−1{\text{37}}{\text{.07 J }}{{\text{K}}^{ - 1}} respectively.
A) −92.843 kJ - 92.843{\text{ kJ}}
B) −47.7723 kJ - 47.7723{\text{ kJ}}
C) −132.5 kJ - 132.5{\text{ kJ}}
D) −176.11 kJ - 176.11{\text{ kJ}}

Explanation

Solution

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 N2{N_2}, H2{H_2} and NH3N{H_3}. Then calculate the value of heat at 50∘C{50^ \circ }{\text{C}}.

Formula Used: CP=(ΔHΔT)P{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:
CP=(ΔHΔT)P{C_P} = {\left( {\dfrac{{\Delta H}}{{\Delta T}}} \right)_P}
Where, CP{C_P} is the molar heat capacity at constant pressure.
ΔH\Delta H is the change in heat.
ΔT\Delta T is the change in temperature.

Rearrange the equation for the change in heat as follows:
ΔH=CP×ΔT\Delta H = {C_P} \times \Delta T
The change in heat is H2−H1{H_2} - {H_1} and the change in temperature is T2−T1{T_2} - {T_1}. Thus,
H2−H1=CP(T2−T1){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:
CP=2×CP(NH3)−[CP(N2)+(3×CP(H2))]{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 P{\text{P}} and 27∘C{27^ \circ }{\text{C}} for N2{N_2}, H2{H_2} and NH3N{H_3} are 28.45{\text{28}}{\text{.45}}, 28.32{\text{28}}{\text{.32}} and 37.07 J K−1{\text{37}}{\text{.07 J }}{{\text{K}}^{ - 1}} respectively. Thus,
⇒CP=2×(37.07 J K−1)−[28.45 J K−1+(3×28.32 J K−1)]\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]
⇒CP=(74.14−113.41) J K−1\Rightarrow {C_P} = \left( {74.14 - 113.41} \right){\text{ J }}{{\text{K}}^{ - 1}}
⇒CP=−39.27 J 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 J K−1=−0.03927 kJ K−1 - 39.27{\text{ J }}{{\text{K}}^{ - 1}} = - 0.03927{\text{ kJ }}{{\text{K}}^{ - 1}}.

Now, calculate the heat of the reaction at 50∘C{50^ \circ }{\text{C}} using equation (1) as follows:
We are given that the heat for the reaction at 27∘C{27^ \circ }{\text{C}} is −91.94 kJ - 91.94{\text{ kJ}}. Thus,
⇒H2−(−91.94 kJ)=−0.03927 kJ K−1(50−27)\Rightarrow {H_2} - \left( { - 91.94{\text{ kJ}}} \right) = - 0.03927{\text{ kJ }}{{\text{K}}^{ - 1}}\left( {50 - 27} \right)
⇒H2+91.94 kJ=−0.03927 kJ K−1(23)\Rightarrow {H_2} + 91.94{\text{ kJ}} = - 0.03927{\text{ kJ }}{{\text{K}}^{ - 1}}\left( {23} \right)
⇒H2=(−0.90321−91.94) kJ\Rightarrow {H_2} = \left( { - 0.90321 - 91.94} \right){\text{ kJ}}
⇒H2=−92.843 kJ\Rightarrow {H_2} = - 92.843{\text{ kJ}}
Thus, the heat of the reaction at 50∘C{50^ \circ }{\text{C}} is −92.843 kJ - 92.843{\text{ kJ}}.

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

Note: We know that CP{C_P} is the molar heat capacity at constant pressure. CP{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 CP{C_P}.