(D) The equilibrium constant expression for the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is given by: $K_C = \frac{[PCl_3][Cl_2]}{[PCl_5]}$ Gi

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

(D) The equilibrium constant expression for the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is given by: $K_C = \frac{[PCl_3][Cl_2]}{[PCl_5]}$ Given $K_C = 1.79$,$[PCl_5] = 1.41 \ M$,and $[PCl_3] = 1.59 \ M$. Substituting these values into the expression: $1.79 = \frac{1.59 \times [Cl_2]}{1.41}$ Solving for $[Cl_2]$: $[Cl_2] = \frac{1.79 \times 1.41}{1.59} = 1.587 \ M$ Rounding to two decimal places,we get $[Cl_2] \approx 1.59 \ M$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

Easy

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. If the equilibrium constant $(K_C)$ for the above reaction at $500 \ K$ is $1.79$ and the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $1.41 \ M$ and $1.59 \ M$,respectively,then the concentration of $Cl_2$ is approximately: (in $M$)

TS EAMCET 2018 All TS EAMCET

A
$1.26$
B
$3.59$
C
$0.59$
D
$1.59$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

6-1.Equilibrium (Chemical Equilibrium)

Subtopic

Law of equilibrium and Equilibrium constant

Previous Year

TS EAMCET 2018 Paper All TS EAMCET years →

For the given reaction,choose the correct expression of $K_{C}$ from the following: $Fe_{(aq)}^{3+} + SCN_{(aq)}^{-} \rightleftharpoons (FeSCN)_{(aq)}^{2+}$

DifficultJEE MAIN 2024

View Solution

The equilibrium constant $(K_c)$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$ at the same temperature is:

MediumAIEEE 2012

View Solution

The following concentrations were obtained for the formation of $NH_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at $500\, K$: $[N_{2}] = 1.5 \times 10^{-2}\, M$,$[H_{2}] = 3.0 \times 10^{-2}\, M$ and $[NH_{3}] = 1.2 \times 10^{-2}\, M$. Calculate the equilibrium constant.

Medium

View Solution

What is the equilibrium constant $(K_C)$ for the given reaction?
$N_2 + O_2 \rightleftharpoons 2 NO$
Where the equilibrium concentrations of $N_2$,$O_2$ and $NO$ are found to be $4 \times 10^{-3} \ M$,$3 \times 10^{-3} \ M$ and $3 \times 10^{-3} \ M$ respectively.

MediumTS EAMCET 2020

View Solution

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$

Medium

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

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. If the equilibrium constant $(K_C)$ for the above reaction at $500 \ K$ is $1.79$ and the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $1.41 \ M$ and $1.59 \ M$,respectively,then the concentration of $Cl_2$ is approximately: (in $M$)

Similar Questions

For the given reaction,choose the correct expression of $K_{C}$ from the following: $Fe_{(aq)}^{3+} + SCN_{(aq)}^{-} \rightleftharpoons (FeSCN)_{(aq)}^{2+}$

The equilibrium constant $(K_c)$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$ at the same temperature is:

The following concentrations were obtained for the formation of $NH_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at $500\, K$: $[N_{2}] = 1.5 \times 10^{-2}\, M$,$[H_{2}] = 3.0 \times 10^{-2}\, M$ and $[NH_{3}] = 1.2 \times 10^{-2}\, M$. Calculate the equilibrium constant.

What is the equilibrium constant $(K_C)$ for the given reaction?$N_2 + O_2 \rightleftharpoons 2 NO$Where the equilibrium concentrations of $N_2$,$O_2$ and $NO$ are found to be $4 \times 10^{-3} \ M$,$3 \times 10^{-3} \ M$ and $3 \times 10^{-3} \ M$ respectively.

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

What is the equilibrium constant $(K_C)$ for the given reaction?
$N_2 + O_2 \rightleftharpoons 2 NO$
Where the equilibrium concentrations of $N_2$,$O_2$ and $NO$ are found to be $4 \times 10^{-3} \ M$,$3 \times 10^{-3} \ M$ and $3 \times 10^{-3} \ M$ respectively.

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$