Question

For which reaction at 297 K, the value of $\frac{K_P}{K_C}$ is maximum and minimum respectively. :-
a)$N_2{O}_4\rightleftharpoons 2{NO}_2$
b)$2S{O}_2+O_2\rightleftharpoons 2{SO}_3$
c)$X+Y\rightleftharpoons 4Z$
d)$A+3B\rightleftharpoons 7C$

• A. d,c
• B. d,b
• C. c,b
• D. d,a

Answer 1

The correct option is B d,b

Given are four equations in the question.

To find: For which reaction $\frac{K_P}{K_C}$ is minimum and maximum respectively

Solution: We know that, $K_P=K_C(RT{)}^{\Delta n_g}$

Where R is gas constant, T is temperature, $\Delta n_g$is change in gaseous moles of reactant and product

$\Rightarrow \frac{K_P}{K_C}=(RT{)}^{\Delta n_g}$

Since we want $\frac{K_P}{K_C}$ in each cases, we will compare $(RT{)}^{\Delta n_g}$ in each case.

Since R is a constant and T is same in each case, hence we will compare $\Delta n_g$ in each case.

Therefore in (a), $\Delta n_g=2-0=2$

In (b), $\Delta n_g=2-(2+1)=-1$

In (c), $\Delta n_g=4-(1+1)=2$

In (d), $\Delta n_g=7-(3+1)=3$

Since , in (b) $\Delta n_g=-1\therefore (RT{)}^{\Delta n_g}$ is minimum

In (d) $\Delta n_g=3\therefore (RT{)}^{\Delta n_g}$ is maximum

Hence, option (B) is correct