JEE PYQ: Chemical Equilibrium Question 24
Question 24 - 2019 (10 Jan Shift 1)
The values of $K_p/K_c$ for the following reactions at 300 K are, respectively: (At 300 K, $RT = 24.62$ dm$^3$ atm mol$^{-1}$)
$\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$; $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$; $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$
(1) 1, 24.62 dm$^3$ atm mol$^{-1}$, 606.0 dm$^6$ atm$^2$ mol$^{-2}$
(2) 1, 24.62 dm$^3$ atm mol$^{-1}$, $1.65 \times 10^{-3}$ dm$^{-6}$ atm$^{-2}$ mol$^2$
(3) 1, 4.1 $\times$ 10$^{-2}$ dm$^{-3}$ atm$^{-1}$ mol, 606 dm$^6$ atm$^2$ mol$^{-2}$
(4) 24.62 dm$^3$ atm mol$^{-1}$, 606.0 dm$^6$ atm$^2$ mol$^{-2}$, $1.65 \times 10^{-3}$ dm$^{-6}$ atm$^{-2}$ mol$^2$
Show Answer
Answer: (2) 1, 24.62 dm$^3$ atm mol$^{-1}$, $1.65 \times 10^{-3}$ dm$^{-6}$ atm$^{-2}$ mol$^2$
Solution
$K_p/K_c = (RT)^{\Delta n_g}$. For $\Delta n_g = 0, 1, -2$: values are 1, 24.62, $1.65 \times 10^{-3}$.
BETA
