The values of K_p/K_c for the following react | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$,which implies $K_p/K_c = (RT)^{\Delta n_g}$.Given $RT = 24.62 \

Vedclass · Accepted Answer

$1, 24.62 \ dm^3 \ atm \ mol^{-1}, 1.65 	imes 10^{-3} \ dm^{-6} \ atm^{-2} \ mol^2$

Vedclass · Answer

(B) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$,which implies $K_p/K_c = (RT)^{\Delta n_g}$.Given $RT = 24.62 \ dm^3 \ atm \ mol^{-1}$.For reaction $(i): N_{2(g)} + O_{2(g)} ightleftharpoons 2NO_{(g)}$,$\Delta n_g = 2 - (1+1) = 0$. So,$K_p/K_c = (RT)^0 = 1$.For reaction $(ii): N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$,$\Delta n_g = 2 - 1 = 1$. So,$K_p/K_c = (RT)^1 = 24.62 \ dm^3 \ atm \ mol^{-1}$.For reaction $(iii): N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$,$\Delta n_g = 2 - (1+3) = -2$. So,$K_p/K_c = (RT)^{-2} = (24.62)^{-2} \approx 1.65 	imes 10^{-3} \ dm^{-6} \ atm^{-2} \ mol^2$.

Similar Questions

The equilibrium constant $K_p$ for the following reaction at $191\,^{\circ}C$ is $1.24$. What is the value of $K_c$? $B_{(s)} + \frac{3}{2}F_{2(g)} \rightleftharpoons BF_{3(g)}$

The equilibrium constant $(K_p)$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{-4}$ and at $127^{\circ} C$ is $0.60 \times 10^{-4}$. Calculate the mean heat of formation of ammonia per mole in this temperature range. (in $cal$)

For the reaction system $A_{(g)} \rightleftharpoons B_{(g)} + 2C_{(g)}$,if the equilibrium concentration of $B$ is doubled,the equilibrium concentration of $C$ will become ....

For the reaction $PCl_5 \, (g) \rightleftharpoons PCl_3 \, (g) + Cl_2 \, (g)$,the two reaction constants $K_p$ and $K_c$ are related to each other by the expression

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The values of $K_p/K_c$ for the following reactions at $300 \ K$ are respectively (At $300 \ K, RT = 24.62 \ dm^3 \ atm \ mol^{-1}$):$(i) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$$(ii) \ N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$$(iii) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$