
The yield of ${\text{N}}{{\text{H}}_{\text{3}}}$in the reaction ${{\text{N}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{ }} \rightleftharpoons {\text{ 2N}}{{\text{H}}_{\text{3}}};{\text{ }}\Delta H{\text{ = - 22}}{\text{.08kcal }}$is affected by:
A. change in pressure and temperature
B. change in temperature and concentration of ${{\text{N}}_{\text{2}}}$
C. change in pressure and concentration of ${{\text{N}}_{\text{2}}}$
D. change in pressure, temperature and concentration of ${{\text{N}}_{\text{2}}}$
Answer
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Hint: Le Chatelier’s principle states that when a system in equilibrium is disturbed by changing the concentrations, changing pressure, changing the temperature, the system shifts its equilibrium, to balance the change of variables.
Complete step by step answer:
The given chemical reaction is
${{\text{N}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{ }} \rightleftharpoons {\text{ 2N}}{{\text{H}}_{\text{3}}};{\text{ }}\Delta H{\text{ = - 22}}{\text{.08kcal }}$
According to Le Chatelier’s principle, there are various factors like a concentration of reactant and product, pressure and temperature affects the position of equilibrium and thus the yield of the product.
The effect of change in concentration of ${{\text{N}}_{\text{2}}}$ on yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ is as follows:
An increase in the concentration of the reactant shifts the equilibrium to the product side while a decrease in the concentration of the reactant shifts the equilibrium to the reactant side.
Similarly, an increase in the concentration of products shifts the equilibrium to the reactant side while a decrease in the concentration of products shifts the equilibrium to the product side.
In the reaction given to us, ${{\text{N}}_{\text{2}}}$ is a reactant, so a change in concentration of ${{\text{N}}_{\text{2}}}$ affects the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$. It will increase or decrease due to an increase or decrease in the concentration of ${{\text{N}}_{\text{2}}}$.
The effect of change in temperature on the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ is as follows:
The negative value of enthalpy of reaction ($\Delta H{\text{ = - 22}}{\text{.08kcal }}$$\Delta H$) indicates that this is an exothermic reaction. Reactions that evolve heat energy are known as exothermic reactions.
In the case of an exothermic reaction, an increase in temperature shifts the equilibrium to the reactant side while a decrease in temperature shifts the equilibrium to the product side. Thus, an increase in temperature will decrease the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ while a decrease in temperature will increase the yield of${\text{N}}{{\text{H}}_{\text{3}}}$.
Thus, a change in temperature affects the yield of${\text{N}}{{\text{H}}_{\text{3}}}$.
Effect of change in pressure on yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ is as follows:
According to Le Chatelier’s principle, an increase in pressure by decreasing volume, shifts the equilibrium to the side of fewer numbers of moles of gas. While a decrease in pressure by increasing volume shifts the equilibrium to the side of greater numbers of moles of gas.
${{\text{N}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{ }} \rightleftharpoons {\text{ 2N}}{{\text{H}}_{\text{3}}};{\text{ }}\Delta H{\text{ = - 22}}{\text{.08kcal }}$
$4{\text{ moles 2 moles}}$
Here, moles of gaseous reactants are greater than moles of a gaseous product. So an increase in the pressure by decreasing volume will increase the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ while a decrease in the pressure by increasing volume will decrease the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$.
Thus, a change in pressure affects the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$.
Hence, the correct answer is option (D) .
Note:
The system where the rate of forward reaction and the backward reaction is equal is said to be at a state of equilibrium. Change in reaction condition that changes in temperature, pressure and concentration changes the position of equilibrium and affects the yield of the product.
Complete step by step answer:
The given chemical reaction is
${{\text{N}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{ }} \rightleftharpoons {\text{ 2N}}{{\text{H}}_{\text{3}}};{\text{ }}\Delta H{\text{ = - 22}}{\text{.08kcal }}$
According to Le Chatelier’s principle, there are various factors like a concentration of reactant and product, pressure and temperature affects the position of equilibrium and thus the yield of the product.
The effect of change in concentration of ${{\text{N}}_{\text{2}}}$ on yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ is as follows:
An increase in the concentration of the reactant shifts the equilibrium to the product side while a decrease in the concentration of the reactant shifts the equilibrium to the reactant side.
Similarly, an increase in the concentration of products shifts the equilibrium to the reactant side while a decrease in the concentration of products shifts the equilibrium to the product side.
In the reaction given to us, ${{\text{N}}_{\text{2}}}$ is a reactant, so a change in concentration of ${{\text{N}}_{\text{2}}}$ affects the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$. It will increase or decrease due to an increase or decrease in the concentration of ${{\text{N}}_{\text{2}}}$.
The effect of change in temperature on the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ is as follows:
The negative value of enthalpy of reaction ($\Delta H{\text{ = - 22}}{\text{.08kcal }}$$\Delta H$) indicates that this is an exothermic reaction. Reactions that evolve heat energy are known as exothermic reactions.
In the case of an exothermic reaction, an increase in temperature shifts the equilibrium to the reactant side while a decrease in temperature shifts the equilibrium to the product side. Thus, an increase in temperature will decrease the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ while a decrease in temperature will increase the yield of${\text{N}}{{\text{H}}_{\text{3}}}$.
Thus, a change in temperature affects the yield of${\text{N}}{{\text{H}}_{\text{3}}}$.
Effect of change in pressure on yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ is as follows:
According to Le Chatelier’s principle, an increase in pressure by decreasing volume, shifts the equilibrium to the side of fewer numbers of moles of gas. While a decrease in pressure by increasing volume shifts the equilibrium to the side of greater numbers of moles of gas.
${{\text{N}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{ }} \rightleftharpoons {\text{ 2N}}{{\text{H}}_{\text{3}}};{\text{ }}\Delta H{\text{ = - 22}}{\text{.08kcal }}$
$4{\text{ moles 2 moles}}$
Here, moles of gaseous reactants are greater than moles of a gaseous product. So an increase in the pressure by decreasing volume will increase the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$ while a decrease in the pressure by increasing volume will decrease the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$.
Thus, a change in pressure affects the yield of ${\text{N}}{{\text{H}}_{\text{3}}}$.
Hence, the correct answer is option (D) .
Note:
The system where the rate of forward reaction and the backward reaction is equal is said to be at a state of equilibrium. Change in reaction condition that changes in temperature, pressure and concentration changes the position of equilibrium and affects the yield of the product.
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