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# In Haber’s process of ammonia manufacture:${{{N}}_{{2}}}{{(g) + 3}}{{{H}}_{{2}}}{{(g) }} \to {{2N}}{{{H}}_{{3}}}{{(g)}}$ ; ${{\Delta }}{{{H}}^{{0}}}_{{{2}}{{{5}}^{{0}}}{{C}}}{{ = - 92}}{{.2 kJ}}$If $C_p$ is independent of temperature, then reaction at ${100^0}{{C}}$ as compared to that of ${{2}}{{{5}}^{{0}}}{{C}}$ will be.A.More endothermicB.Less endothermicC.More exothermicD.Less exothermic

Last updated date: 20th Jun 2024
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Hint:The Nitrogen from the air is combined with hydrogen from methane to form ammonia. This is a reversible reaction. It is the industrial method of preparing ammonia. High temperature and pressure should be maintained in this process. This reaction is an exothermic process that involves the release of energy. The nitrogen is obtained by liquefaction of air and hydrogen is obtained by steam reforming of natural gas. ${{{C}}_{{p}}}$ is the specific heat capacity at constant pressure.

Standard enthalpy of formation of ammonia at ${{2}}{{{5}}^{{0}}}{{C}}$ its given as ${{ - 92}}{{.2 kJ}}$ which shows this is an exothermic reaction
Here, ${{{C}}_{{p}}}$ each molecule is given and we can calculate the change in heat capacity.
${{\Delta }}{{{C}}_{{p}}}{{ = 2}}{{{C}}_{{{p(N}}{{{H}}_{{3}}}{{)}}}}{{ - }}{{{C}}_{{{p(}}{{{N}}_{{2}}}{{)}}}}{{ - 3}}{{{C}}_{{{P(}}{{{H}}_{{2}}}{{)}}}}$
$\Rightarrow {{\Delta }}{{{C}}_{{P}}}{{ = 2(35}}{{.1) - (29}}{{.1) - 3(28}}{{.8) J}}{{{K}}^{{{ - 1}}}}{{mo}}{{{l}}^{{{ - 1}}}}$
$\Rightarrow {{\Delta }}{{{C}}_{{P}}}{{ = - 45}}{{.3 J}}{{{K}}^{{{ - 1}}}}{{mo}}{{{l}}^{{{ - 1}}}}$
We know that at constant pressure,
The change in enthalpy is the difference between the enthalpy of the product and the enthalpy of the reactant.
${{\Delta H = }}{{{H}}_{{p}}}{{ - }}{{{H}}_{{R}}} \to eq1$
We know that ${{H = }}{{{C}}_{{p}}}{{\Delta T}}$ at constant pressure,
So we can write $\Delta {{H = \Delta }}{{{C}}_{{p}}}{{dT}} \to {{eq2}}$
From equation 1 and 2 we can write as ${{\Delta }}{{{H}}_2}{{ - \Delta }}{{{H}}_1}{{ = \Delta }}{{{C}}_{{p}}}{{dT}}$
Here, ${{\Delta }}{{{H}}_{{2}}}$ is the enthalpy at ${100^0}{{C}}$ and ${{\Delta }}{{{H}}_{{1}}}$is the enthalpy change at ${{2}}{{{5}}^{{0}}}{{C}}$
So it can be written as ${{\Delta }}{{{H}}_{{2}}}{{ - ( - 92}}{{.2 \times 1}}{{{0}}^{{3}}}{{) = - 45}}{{.3(}}{{{T}}_{{2}}}{{ - }}{{{T}}_{{1}}}{{)}}$
$\Rightarrow {{\Delta }}{{{H}}_{{2}}}{{ - ( - 92}}{{.2 \times 1}}{{{0}}^{{3}}}{{) = - 45}}{{.3(100 - 25)}}$
$\Rightarrow$ ${{\Delta }}{{{H}}_{{2}}}{{ = - 95600 J/mol}}$
$\Rightarrow$ ${{\Delta }}{{{H}}_{{2}}}{{ = - 95}}{{.6 kJ/mol}}$
It is less than the enthalpy change at ${{2}}{{{5}}^{{0}}}{{C}}$ it is more negative, so we can say that it is more exothermic.
The correct answer is option C.

Note:
Kirchhoff’s law gives a relation between the temperature dependence of thermal quantities in a reaction by the difference in heat capacities of the products and reactants. At constant pressure, the heat capacity is the enthalpy divided by change in temperature. Exothermic reactions have enthalpy value always negative whereas for endothermic, the value is positive.