$5.1 \ g$ $NH_4SH$ 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 $N = 14 \ g \ mol^{-1}$)
- A$0.242 \times 10^{-4} \ atm^2$
- B$1 \times 10^{-4} \ atm^2$
- C$4.9 \times 10^{-3} \ atm^2$
- D$0.242 \ atm^2$
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Similar Questions
For the formation of ammonia from its constituent elements ($1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$) in a closed vessel of volume $V \ L$,the value of $K_C$ is [units of $K_C = mol^{-2} \ L^2$].
If the equilibrium constants of the following equilibria $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$ and $2 SO_3 \rightleftharpoons 2 SO_2 + O_2$ are given by $K_1$ and $K_2$ respectively, which of the following relations is correct?
For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_p = 41$ at $400 \ K$. Calculate $K_c$ for the following reactions at $400 \ K$:
$(a)$ $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$
$(b)$ $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$(c)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$
$(a)$ $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$
$(b)$ $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$(c)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$
Difficult
View SolutionFor which of the following reactions is the relation $\frac{K_p}{K_c} + \log(RT) = 0$ correct?
Medium
View SolutionWhat is the unit of $K_p$ for the reaction?
$CS_{2(g)} + 4H_{2(g)} \to CH_{4(g)} + 2H_2S_{(g)}$
$CS_{2(g)} + 4H_{2(g)} \to CH_{4(g)} + 2H_2S_{(g)}$
Medium
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