Question

At $1000K$, from the data :
${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$; $\Delta H = - 123.77kJmo{l^{ - {\mathbf{1}}}}$

Substance	${N_2}$	${H_2}$	$N{H_3}$
P/R	$3.5$	$3.5$	$4$

Calculate the heat of formation of $N{H_3}$ ​ at $300K$.
A.$- 44.42kJmo{l^{ - 1}}$
B.$- 88.85kJmo{l^{ - 1}}$
C.$+ 44.42kJmo{l^{ - 1}}$
D.$+ 88.85kJmo{l^{ - 1}}$

Answer 1

We need to know the effect of Temperature on Heat of Reaction. Chemical or physical processes where heat change takes place, generally depends on the temperature at which the process takes place. This dependence is given by the Kirchhoff equation. This dependence is mathematically expressed in the form of what is known as Kirchhoff equation after G. R. Kirchhoff (1858) who first developed this equation. The equation may easily be derived with the help of the first law of thermodynamics.

Complete answer:
We need to know that the Kirchhoff equation relates the heat of reaction with the definite heats of a structure before and after the reaction.
Consider the process in which the reactants in state A at temperature ${T_1}$ are converted into products in state B a temperature ${T_2}$.Then The reactants in state A at temperature ${T_1}$ are heated to a temperature ${T_2}$. The heat absorbed is $\left( {\Delta T} \right){\left( {{C_P}} \right)_A}$. Where, $\Delta T = {T_2}-{\text{ }}{T_1}$ , and ${\left( {{C_P}} \right)_A}$ is the heat capacity of the reactants in the state A. The reaction is now allowed to take place at this temperature and the heat change for the process is ${\left( {{H_B}-{H_A}} \right)_2} = \Delta {H_2}$
The total heat change for the process = $\left( {\Delta T} \right){\left( {{C_P}} \right)_A} + \Delta {H_2}$
Also, the reactants in state A at temperature ${T_1}$ are considered to products in state B at the same temperature. The heat change =${\left( {{H_B}-{H_A}} \right)_1} = \Delta {H_1}$. The temperature of the products is then raised from ${T_1}$to ${T_2}$ and the heat absorbed is $\left( {\Delta T} \right){\left( {{C_P}} \right)_B}$, where ${\left( {{C_P}} \right)_B}$-the heat capacity of the products.
The total heat change for the process $= \left( {\Delta T} \right){\left( {{C_P}} \right)_B} + \Delta {H_1}$
According to Hess’s Law of Constant Heat summation, if the reaction takes place in several steps then its Total heat change is the sum of the heat changes of the individual reactions.
Therefore, $\left( {\Delta T} \right){\left( {{C_P}} \right)_A} + \Delta {H_2} = \left( {\Delta T} \right){\left( {{C_P}} \right)_B} + \Delta {H_1}$
Or , $\Delta {H_2} - \Delta {H_1} = [{({C_P})_B}-{({C_P})_A}]x\left( {\Delta T} \right)$
$\Delta {H_2} - \Delta {H_1} = \Delta {C_P}\left( {\Delta T} \right)$
As given in the question,
$\Delta {H_2}\left( {1000K} \right) = - 123.77kJ/mol$
$\Delta T = 1000K - 300K$
On simplification we get,
$\Rightarrow \Delta T = 700K$
$\Delta {C_P} = 2{C_P}(N{H_3}) - [{C_P}({N_2}) + 3{C_P}({H_2})]$
$\Rightarrow \Delta {C_p} = - 6 \times 8.314 \times {10^{ - 3}}kJ$
$\Delta {H_1}\left( {300K} \right) = ?$
From Kirchhoff's equation.
$\Delta {H_2}\left( {1000K} \right) = \Delta {H_1}\left( {300K} \right) + \Delta Cp\Delta T$
On substituting the known values we get,
$\Rightarrow - 123.77 = \Delta {H_1}\left( {300K} \right) + ( - 6 \times 8.314 \times {10^{ - 3}}) \times 700$
On simplification we get,
$\Rightarrow \Delta {H_1}\left( {300K} \right) = - 88.85kJ$
This value implies for $2$ moles of $N{H_3}$
Therefore, heat of formation of $N{H_3}$ ​ at $300K = \dfrac{{ - 88.85}}{2}$
$= - 44.42kJmo{l^{ - {\mathbf{1}}}}$

Note:
It must be noted that Heat formation or heat change is also known as standard enthalpy change and is independent of the path between initial state (reactants) and final state (products). The combination of Kirchoff’s Law and Hess’s Law can be used to calculate the formation of heat at different temperatures.