Consider the equilibrium,$H_2 + I_2 \rightleftharpoons 2 HI$. Calculate the equilibrium constant of the reverse reaction when the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $1.14 \times 10^{-2} \ mol \ L^{-1}$,$0.12 \times 10^{-2} \ mol \ L^{-1}$,and $2.52 \times 10^{-2} \ mol \ L^{-1}$,respectively.
- A$46.4$
- B$0.021$
- C$18.42$
- D$0.054$
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$5.1 \ g$ $NH_4SH$ is introduced in a $3.0 \ L$ evacuated flask at $327 \ ^\circ C$. $30\%$ of the solid $NH_4SH$ decomposes into $NH_3$ and $H_2S$ gases. The $K_p$ of the reaction at $327 \ ^\circ C$ is ($R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,molar mass of $S = 32 \ g \ mol^{-1}$,molar mass of $N = 14 \ g \ mol^{-1}$)
DifficultJEE MAIN 2019
View SolutionFor the reaction: $Fe_{2}N_{(s)} + \frac{3}{2}H_{2(g)} \rightleftharpoons 2Fe_{(s)} + NH_{3(g)}$,which of the following relations is correct?
DifficultJEE MAIN 2020
View SolutionOne mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:
$2 \ mol$ $PCl_5$ is heated in a $4 \ L$ closed vessel at a definite temperature. At equilibrium,$55\%$ of $PCl_5$ remains undissociated. Find $K_c$. Reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
Medium
View SolutionThe value of $\log \ K$ for the reaction $A \rightleftharpoons B$ at $298 \ K$ is (Nearest integer).
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)
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