(A) $\text{The dissociation reaction is: } 2HI(g) \rightleftharpoons H_2(g) + I_2(g)$$\text{Initial moles of } HI = 3.2 \text{ mol}$$\text{Degree of d

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

(A) $\text{The dissociation reaction is: } 2HI(g) \rightleftharpoons H_2(g) + I_2(g)$$\text{Initial moles of } HI = 3.2 \text{ mol}$$\text{Degree of dissociation } (\alpha) = 22\% = 0.22$$\text{Moles of } HI \text{ dissociated} = \text{Initial moles} \times \alpha = 3.2 \times 0.22 = 0.704 \text{ mol}$$\text{Moles of } HI \text{ remaining at equilibrium} = \text{Initial moles} - \text{Dissociated moles}$$\text{Moles of } HI = 3.2 - 0.704 = 2.496 \text{ mol}$

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Degree of dissociation and Vapour density

Easy

$3.2 \text{ moles of hydrogen iodide are heated in a closed vessel at } 444^{\circ}C \text{ until equilibrium is reached. If the degree of dissociation at this temperature is } 22\%, \text{ then the number of moles of hydrogen iodide remaining at equilibrium will be:}$

A
$2.496$
B
$1.87$
C
$2$
D
$4$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

6-1.Equilibrium (Chemical Equilibrium)

Subtopic

Degree of dissociation and Vapour density

For the reaction $P \rightleftharpoons Q + R$,initially $2 \, \text{mol}$ of $P$ is taken. At equilibrium,$0.5 \, \text{mol}$ of $P$ dissociates. The degree of dissociation will be ...... .

Medium

View Solution

At temperature $T$,the dissociation equilibrium of $AB_{2(g)}$ is given as: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. If the degree of dissociation $x$ is very small compared to $1$,then the expression for the equilibrium constant $K_p$ in terms of $x$ and total pressure $P$ is:

Medium

View Solution

The vapor density of a volatile substance with respect to $CH_4$ $(CH_4 = 1)$ is $4$. What is its molecular weight?

Difficult

View Solution

The density of air is $0.00130 \ g/mL$. The vapour density of air will be

Difficult

View Solution

In a closed vessel,the dissociation of phosphorus pentachloride occurs as: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. If the total pressure at equilibrium is $P$ and the degree of dissociation of $PCl_5$ is $x$,then the partial pressure of $PCl_3$ will be:

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

$3.2 \text{ moles of hydrogen iodide are heated in a closed vessel at } 444^{\circ}C \text{ until equilibrium is reached. If the degree of dissociation at this temperature is } 22\%, \text{ then the number of moles of hydrogen iodide remaining at equilibrium will be:}$

Similar Questions

For the reaction $P \rightleftharpoons Q + R$,initially $2 \, \text{mol}$ of $P$ is taken. At equilibrium,$0.5 \, \text{mol}$ of $P$ dissociates. The degree of dissociation will be ...... .

At temperature $T$,the dissociation equilibrium of $AB_{2(g)}$ is given as: $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. If the degree of dissociation $x$ is very small compared to $1$,then the expression for the equilibrium constant $K_p$ in terms of $x$ and total pressure $P$ is:

The vapor density of a volatile substance with respect to $CH_4$ $(CH_4 = 1)$ is $4$. What is its molecular weight?

The density of air is $0.00130 \ g/mL$. The vapour density of air will be

In a closed vessel,the dissociation of phosphorus pentachloride occurs as: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$. If the total pressure at equilibrium is $P$ and the degree of dissociation of $PCl_5$ is $x$,then the partial pressure of $PCl_3$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module