(A) Initial moles of $NH_4HS = \frac{5.1 \ g}{51 \ g/mol} = 0.1 \ mol$.Reaction: $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$.At equilibri

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(A) Initial moles of $NH_4HS = \frac{5.1 \ g}{51 \ g/mol} = 0.1 \ mol$.Reaction: $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$.At equilibrium,$20\%$ of $0.1 \ mol$ decomposes,so moles of $NH_3 = 0.1 \times 0.2 = 0.02 \ mol$ and moles of $H_2S = 0.02 \ mol$.Volume of vessel $= 2 \ L$,Temperature $T = 27 + 273 = 300 \ K$.Partial pressure of each gas: $P = \frac{nRT}{V} = \frac{0.02 \times 0.082 \times 300}{2} = 0.246 \ atm$.$K_p = P_{NH_3} \times P_{H_2S} = (0.246) \times (0.246) = 0.060516 \approx 6.05 \times 10^{-2}$.Comparing with $x \times 10^{-2}$,we get $x = 6$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

Medium

When $5.1 \ g$ of solid $NH_4HS$ is introduced into a two litre evacuated flask at $27^{\circ}C$,$20\%$ of the solid decomposes into gaseous ammonia and hydrogen sulphide. The $K_p$ for the reaction at $27^{\circ}C$ is $x \times 10^{-2}$.
The value of $x$ is ....... . (Integer answer)
[Given $R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}$]

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A
$6$
B
$60$
C
$0.6$
D
$600$

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6-1.Equilibrium (Chemical Equilibrium)

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Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

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When $1 \ mol$ of $H_2$ and $1 \ mol$ of $N_2$ are enclosed in a $5 \ L$ vessel and the reaction is allowed to attain equilibrium,it is found that at equilibrium there is $x \ mol$ of $H_2$. The number of moles of $NH_3$ would be

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$2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2_{(g)}}$. If $NOBr$ is $40\%$ dissociated at a certain temperature and a total pressure of $0.30 \text{ atm}$,the $K_p$ for the reaction $2NO_{(g)} + Br_{2_{(g)}} \rightleftharpoons 2NOBr_{(g)}$ is:

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the value of $K_c$ at $440 \ ^oC$ is $50$. If the reaction is initiated in a $1 \ L$ flask with $1 \ mol$ of $H_2$,$2 \ mol$ of $I_2$,and $3 \ mol$ of $HI$,then the equilibrium concentration of $HI$ will be .......... $M$.

At $67\,^{\circ}C$ and $1\ bar$ pressure,dinitrogen tetraoxide is $50\%$ dissociated into nitrogen dioxide. $\Delta G^{\circ}$ for the process $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is $(R = \frac{25}{3} \ J \ K^{-1} \ mol^{-1}, \ln 2 = 0.7, \ln 3 = 1.1)$.

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