Question

Consider the following statements.
In the Haber’s method of synthesis of ammonia:
1. An increase of pressure favors the formation of $N{H_3}$.
2. A decrease of pressure produces more $N{H_3}$.
3. An increase of temperature dissociates $N{H_3}$.
4. Addition of inert gas at constant pressure favours the formation of $N{H_3}$.
Which of the statements given above are correct?
A. 1 and 3
B. 2 and 4
C. 1 and 4
D. 2 and 3

Answer 1

Hint: Haber's process is an example of Homogeneous type-III gaseous reaction in which \[\Delta {n_g} < 0\] . Thus, on applying law of mass action on such reactions, we get to know that the dissociation increases with increase in the pressure and decreases with the increase in the volume. And on further increasing the temperature, the ammonia formed will dissociate into nitrogen gas.

Complete step by step answer:
According to the reaction involved in the Haber’s process, we have,
${N_2} + 3{H_2}$ $ \to 2N{H_3}$
Here, $\Delta {n_g} = {n_p} - {n_r} = - ve$
Thus, in the process where the overall stoichiometric coefficients of gaseous mixture are negative, the degree of dissociation(x) is directly proportional to pressure and inversely proportional to volume. In the case of Haber’s process, if the temperature is further increased, it will dissociate the ammonia into nitrogen and other products. On increasing the pressure, dissociation of reactants also increases and thus, with it, the formation of ammonia is also increased.
Hence, the correct option is (A).

Note:
Here, in this question, one must have to understand that the values of $K_c$ and $K_p$ depend upon the value and sign of \[\Delta {n_g} < 0\] and based on that, the equation is marked as type I, II or III. Thus, the law of mass action is helpful to predict the variation of dissociation with the physical parameters such as pressure, volume, mixing of inert gases, etc.