(A) The decomposition reaction is: $NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$.Initially,the pressure of $NH_3$ is $P_{NH_3} = 0.50 \ atm$ and $P

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

(A) The decomposition reaction is: $NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$.Initially,the pressure of $NH_3$ is $P_{NH_3} = 0.50 \ atm$ and $P_{H_2S} = 0 \ atm$.Let the increase in pressure due to the decomposition be $x \ atm$.At equilibrium,$P_{NH_3} = 0.50 + x$ and $P_{H_2S} = x$.The total pressure at equilibrium is $P_{total} = P_{NH_3} + P_{H_2S} = (0.50 + x) + x = 0.84 \ atm$.$0.50 + 2x = 0.84 \implies 2x = 0.34 \implies x = 0.17 \ atm$.At equilibrium,$P_{NH_3} = 0.50 + 0.17 = 0.67 \ atm$ and $P_{H_2S} = 0.17 \ atm$.The equilibrium constant $K_p$ is given by $K_p = P_{NH_3} \times P_{H_2S} = 0.67 \times 0.17 = 0.1139 \approx 0.11$.

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

Medium

Solid $NH_4HS$ is placed in a flask containing $NH_3$ gas at a certain temperature and a pressure of $0.50 \ atm$. The $NH_4HS$ decomposes to form $NH_3$ gas and $H_2S$ gas. When equilibrium is established in the flask,the total pressure increases to $0.84 \ atm$. What is the equilibrium constant $(K_p)$ for the decomposition of $NH_4HS$ at this temperature?

A
$0.11$
B
$0.18$
C
$0.22$
D
$0.33$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

6-1.Equilibrium (Chemical Equilibrium)

Subtopic

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

Consider the following chemical equilibrium of the gas phase reaction at a constant temperature: $A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$. If $p$ is the total pressure,$K_p$ is the equilibrium constant,and $\alpha$ is the degree of dissociation,then which of the following is true at equilibrium?

MediumJEE MAIN 2025

View Solution

In a $0.25 \ L$ tube,$4 \ mol$ of $NO$ undergoes dissociation. If the degree of dissociation is $10\%$,then the value of $K_c$ for the reaction $2NO \rightleftharpoons N_2 + O_2$ will be:

Medium

View Solution

Two equilibria,$AB \rightleftharpoons A^{+} + B^{-}$ and $AB + B^{-} \rightleftharpoons AB_2^-$,are simultaneously maintained in a solution with equilibrium constants $K_1$ and $K_2$ respectively. The ratio of $[A^{+}]$ to $[AB_2^-]$ in the solution is

MediumAIIMS 2016

View Solution

$CH_3COCH_{3(g)} \rightleftharpoons C_2H_{6(g)} + CO_{(g)}$. The initial pressure of $CH_3COCH_3$ is $100 \ mm$. When equilibrium is set up,the mole fraction of $CO_{(g)}$ is $\frac{1}{4}$. Hence,the partial pressure of $CO$ is:

Medium

View Solution

At $783 \, K$ in the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the molar concentrations $(mol \, L^{-1})$ of $H_2, I_2$ and $HI$ at some instant of time are $0.1, 0.2$ and $0.4$,respectively. If the equilibrium constant is $46$ at the same temperature,then as the reaction proceeds:

MediumKVPY 2011

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

In a $0.25 \ L$ tube,$4 \ mol$ of $NO$ undergoes dissociation. If the degree of dissociation is $10\%$,then the value of $K_c$ for the reaction $2NO \rightleftharpoons N_2 + O_2$ will be:

Two equilibria,$AB \rightleftharpoons A^{+} + B^{-}$ and $AB + B^{-} \rightleftharpoons AB_2^-$,are simultaneously maintained in a solution with equilibrium constants $K_1$ and $K_2$ respectively. The ratio of $[A^{+}]$ to $[AB_2^-]$ in the solution is

$CH_3COCH_{3(g)} \rightleftharpoons C_2H_{6(g)} + CO_{(g)}$. The initial pressure of $CH_3COCH_3$ is $100 \ mm$. When equilibrium is set up,the mole fraction of $CO_{(g)}$ is $\frac{1}{4}$. Hence,the partial pressure of $CO$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module