The standard equilibrium constant,K_p at 298 | vedclass

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

Question

(C) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftha

Vedclass · Accepted Answer

$3.56 	imes 10^8$

Vedclass · Answer

(C) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.For the reaction $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - (1 + 3) = -2$.Given $K_p = 5.8 	imes 10^5$,$T = 298 \, K$,and $R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1}$.Rearranging the formula to solve for $K_c$: $K_c = K_p(RT)^{-\Delta n_g} = K_p(RT)^2$.Substituting the values: $K_c = (5.8 	imes 10^5) 	imes (0.08314 	imes 298)^2$.$K_c = (5.8 	imes 10^5) 	imes (24.77572)^2$.$K_c = (5.8 	imes 10^5) 	imes 613.836$.$K_c \approx 3.56 	imes 10^8$.

Similar Questions

The equilibrium $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ shows that $K_P$ (in $atm$) is double the value of $K_C$ (in $mol/L$) at a particular temperature $T$. Then,$T$ is $...... \ K$.

For the reaction $2HI \rightleftharpoons H_2 + I_2$,which of the following is true?

At $1100 \ K$ temperature,$CaCO_{3(s)}$ and $CaO_{(s)}$ are in equilibrium. The pressure of $CO_{2(g)}$ is $2.0 \times 10^5 \ Pa$.
Find the equilibrium constant $(K_p)$ for the reaction: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

For the reaction,$PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the value of $K_c$ at $250 \ ^oC$ is $26$. The value of $K_p$ at this temperature will be:

Assertion : For reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the unit of $K_C$ is $L^2 \, mol^{-2}$.
Reason : For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_C = \frac{[NH_3]^2}{[N_2][H_2]^3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module