Question

In Haber’s process of ammonia manufacture:
${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)$ , $\Delta {H^ \circ }_{{{25}^ \circ }C} = - 92.2kJ$

Molecules	${N_2}(g)$	${H_2}(g)$	$N{H_3}\left( g \right)$
${C_r}J{K^{ - 1}}mo{l^{ - 1}}$	$29.1$	$28.8$	$35.1$

If ${C_P}$ is independent of temperature, then reaction at ${100^ \circ }C$ compared to that of ${25^ \circ }C$ will be:
A. more endothermic
B. less endothermic
C. more exothermic
D. less exothermic

Answer 1

To solve this equation, we need to first understand Kirchhoff's equation. From Kirchhoff's law we can describe that the enthalpy of a reaction varies with change in temperature. We can also say that the enthalpy of any substance increases with temperature, which means both the products and the reactants' enthalpies increase. So, the overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different.

Complete answer:
We have to know that an exothermic reaction is a process that releases energy from the system to its surroundings whereas an endothermic reaction is just the vice versa. According to Kirchhoff's law we say that at constant pressure, the heat capacity is equal to change in enthalpy divided by the change in temperature. So, the formula of heat capacity is written as:
${C_P} = \dfrac{{\Delta H}}{{\Delta T}}$
The enthalpy change can be written as:
$\Delta {H_2} = \Delta {H_1} + \Delta {C_P}({T_2} - {T_1})$
Now we find the T1 value as,
${T_1} = {25^ \circ }C = (25 + 273)K = 288K$
Now we find the T2 value as,
${T_2} = {100^ \circ }C = (100 + 273)K = 373K$
Given data,
$\Delta {H_1} = - 92.2 \times {10^3}J$
$\Delta {C_P} = 2{C_P}(N{H_3}) - [{C_P}({N_2}) + 3{C_P}({H_2})]$
On substituting the known values we get,
$\Rightarrow \Delta {C_P} = 2 \times 35.1 - [29.1 + 3 \times 28.8]J{K^{ - 1}}$
On simplification we get,
$\Delta {C_P} = - 45.3J{K^{ - 1}}$
Now we substitute the known values in enthalpy change formula we get,
$\Rightarrow \Delta {H_2} = - 92.2 \times {10^3}J + ( - 45.3)(373 - 288)$
$\Rightarrow \Delta {H_2} = - 92.2 \times {10^3} - 3850.5$
On simplification we get,
$\Rightarrow \Delta {H_2} = - 96.050KJ$
Therefore, we can say that the reaction occurring at $100^\circ C$ is less exothermic compared to the reaction occurring at $25^\circ C$ . Therefore, the option D is correct.
As the enthalpy change at $100^\circ C$ is less than the value of enthalpy change at $25^\circ C$ .
$\Delta {H_2} < \Delta {H_1}$
$- 96.050KJ < \- 92.2 \times {10^3}J$ .
Therefore, the option D is correct.

Note:
After we have solved this question, we also need to understand the basic difference between an exothermic and endothermic reaction. Endothermic and exothermic reactions both are chemical reactions that have the capability to absorb and release heat, respectively. For example, photosynthesis can be seen in an endothermic reaction. Also, combustion is an example of an exothermic reaction. The main categorization of the reaction as endothermic and exothermic reaction depends on the net heat transfer of the reaction.