In a one-litre flask,6 moles of A undergoes t | vedclass

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

Question

(C) At $T_1 	ext{ K}$: $A_{(g)} ightleftharpoons P_{(g)}$$t=0$: $6 \quad 0$$t=\infty$: $6-4=2 \quad 4$ (from plot)$\Rightarrow K_{c_1} = \frac{4}{2

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) At $T_1 	ext{ K}$: $A_{(g)} ightleftharpoons P_{(g)}$$t=0$: $6 \quad 0$$t=\infty$: $6-4=2 \quad 4$ (from plot)$\Rightarrow K_{c_1} = \frac{4}{2} = 2$At $T_2 	ext{ K}$: $A_{(g)} ightleftharpoons P_{(g)}$$t=0$: $6 \quad 0$$t=\infty$: $6-2=4 \quad 2$ (from plot)$\Rightarrow K_{c_2} = \frac{2}{4} = \frac{1}{2}$Since $\Delta n_g = 0$,$K_p = K_c$.$\Delta G_2^{\Theta} = -RT_2 \ln K_{p_2} = -RT_2 \ln \frac{1}{2} = RT_2 \ln 2$$\Delta G_1^{\Theta} = -RT_1 \ln K_{p_1} = -R(2T_2) \ln 2 = -2RT_2 \ln 2$$\Delta G_2^{\Theta} - \Delta G_1^{\Theta} = RT_2 \ln 2 - (-2RT_2 \ln 2) = 3RT_2 \ln 2 = RT_2 \ln 2^3 = RT_2 \ln 8$Given $\Delta G_2^{\Theta} - \Delta G_1^{\Theta} = RT_2 \ln x$,therefore $x = 8$.

Similar Questions

For the gas phase reaction; $C_2H_{4(g)} + H_{2(g)} \rightleftharpoons C_2H_{6(g)}$,$\Delta H = -32.7 \ kcal$ carried out in a vessel,the equilibrium concentration of $C_2H_4$ can be increased by:

The reaction quotient $Q_{C}$ is useful in predicting the direction of the reaction. Which of the following is incorrect?

$A$ vessel at $1000 \ K$ contains $CO_2$ with a pressure of $0.5 \ atm$. Some of $CO_2$ is converted into $CO$ on addition of graphite. If total pressure at equilibrium is $0.8 \ atm$,then $K_P$ is : (in $atm$)

In a $2 \ L$ vessel,$1 \ mol$ of $H_2$ and $2 \ mol$ of $I_2$ are taken. If at equilibrium the moles of $H_2$ are $0.2$,then the moles of $I_2$ and $HI$ at equilibrium will be respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module