
Pure ammonia is placed in a vessel at a temperature where its dissociation constant \[\left( \alpha \right)\] is appreciable. At equilibrium:
A.\[{K_p}\] does not change significantly with pressure
B.\[\alpha \] does not change with pressure
C.The concentration of \[N{H_3}\] does not change with pressure
D.The concentration of hydrogen is less than nitrogen
Answer
497.4k+ views
Hint: The term equilibrium refers to the state where the concentration of both reactants and products are kept constant. The equilibrium constant is indicated by K. Dissociation constant is something which measures the extent of dissociation which means the separation of particles. It is indicated by \[\alpha \] .
Complete answer:
In the question, it is mentioned that the pure ammonia in the vessel is undergoing dissociation, but the dissociation constant is not mentioned. Let’s write the equation and equilibrium constant for it.
\[2N{H_3} \rightleftharpoons {N_2} + 3{H_2}\]
\[ {K_c} = {\dfrac{{\left[ {{N_2}} \right]\left[ {{H_2}} \right]}}{{{{\left[ {N{H_3}} \right]}^2}}}^3}\]
Here [] indicates the concentration. Now, let us check the options one after the other.
Option A: It is mentioned that \[{K_p}\] does not change with pressure which is true because it only changes with temperature and remains unchanged with other factors like pressure, volume etc. Hence, it is correct option
Option B: A. In the above reaction, according to Le Chateleir’s principle, increase in pressure shifts the equilibrium towards left and decrease in pressure shifts the equilibrium towards right. Hence \[\alpha \] changes with change in pressure. Hence, it is incorrect option
Option C: With change in pressure, \[\alpha \] changes and hence the concentration also changes because \[\alpha = C{K_a}\] . Here \[\alpha \] is directly related to K which is in turn dependent on the concentration. So, this is an incorrect option.
Option D: The concentration of hydrogen is more when compared to nitrogen as there are 3 moles of hydrogen and only 2 moles of nitrogen. So, it is incorrect.
By the above explanation, the correct option is A.
Note:
I mentioned that equilibrium constant K changes only with the temperature. If it is an endothermic reaction then with increase in temperature, K value increases and if it is an exothermic reaction then with increase in temperature, K value decreases and vice versa.
Complete answer:
In the question, it is mentioned that the pure ammonia in the vessel is undergoing dissociation, but the dissociation constant is not mentioned. Let’s write the equation and equilibrium constant for it.
\[2N{H_3} \rightleftharpoons {N_2} + 3{H_2}\]
\[ {K_c} = {\dfrac{{\left[ {{N_2}} \right]\left[ {{H_2}} \right]}}{{{{\left[ {N{H_3}} \right]}^2}}}^3}\]
Here [] indicates the concentration. Now, let us check the options one after the other.
Option A: It is mentioned that \[{K_p}\] does not change with pressure which is true because it only changes with temperature and remains unchanged with other factors like pressure, volume etc. Hence, it is correct option
Option B: A. In the above reaction, according to Le Chateleir’s principle, increase in pressure shifts the equilibrium towards left and decrease in pressure shifts the equilibrium towards right. Hence \[\alpha \] changes with change in pressure. Hence, it is incorrect option
Option C: With change in pressure, \[\alpha \] changes and hence the concentration also changes because \[\alpha = C{K_a}\] . Here \[\alpha \] is directly related to K which is in turn dependent on the concentration. So, this is an incorrect option.
Option D: The concentration of hydrogen is more when compared to nitrogen as there are 3 moles of hydrogen and only 2 moles of nitrogen. So, it is incorrect.
By the above explanation, the correct option is A.
Note:
I mentioned that equilibrium constant K changes only with the temperature. If it is an endothermic reaction then with increase in temperature, K value increases and if it is an exothermic reaction then with increase in temperature, K value decreases and vice versa.
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