Difficult
Which of the following reactions will be affected by increasing the pressure? Also,mention whether the change will cause the reaction to proceed in the forward or backward direction.
$(i)$ $CaCl_{2(s)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$
$(ii)$ $CH_{4(g)} + 2S_{2(g)} \rightleftharpoons CS_{2(g)} + 2H_{2}S_{(g)}$
$(iii)$ $CO_{2(g)} \rightleftharpoons C_{(s)} + 2CO_{(g)}$
$(iv)$ $2H_{2(g)} + CO_{(g)} \rightleftharpoons CH_{3}OH_{(g)}$
$(v)$ $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
$(vi)$ $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_{2}O_{(g)}$
$(i)$ $CaCl_{2(s)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$
$(ii)$ $CH_{4(g)} + 2S_{2(g)} \rightleftharpoons CS_{2(g)} + 2H_{2}S_{(g)}$
$(iii)$ $CO_{2(g)} \rightleftharpoons C_{(s)} + 2CO_{(g)}$
$(iv)$ $2H_{2(g)} + CO_{(g)} \rightleftharpoons CH_{3}OH_{(g)}$
$(v)$ $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
$(vi)$ $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_{2}O_{(g)}$
(A) In chemical equilibrium,the effect of pressure is observed only if the change in the number of gaseous moles,$\Delta n_g \neq 0$.
According to Le Chatelier's principle,increasing the pressure shifts the equilibrium in the direction where the number of gaseous moles decreases.
According to Le Chatelier's principle,increasing the pressure shifts the equilibrium in the direction where the number of gaseous moles decreases.
| $Reaction$ | $Effect$ $of$ $increasing$ $pressure$ |
| $(i).$ $\Delta n_g = +1$ | Backward $direction$ |
| $(ii).$ $\Delta n_g = 0$ | No $effect$ |
| $(iii).$ $\Delta n_g = +1$ | Backward $direction$ |
| $(iv).$ $\Delta n_g = -2$ | Forward $direction$ |
| $(v).$ $\Delta n_g = +1$ | Backward $direction$ |
| $(vi).$ $\Delta n_g = +1$ | Backward $direction$ |
Explore More
Similar Questions
For the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$,$\Delta H^o = -198 \, kJ$. According to Le Chatelier's principle,the favorable conditions for the forward reaction are:
Easy
View SolutionIn which of the following reactions at equilibrium,the position of the equilibrium shifts towards the products,if the total pressure is increased?
$(I)$ $X_{2(g)} + 3Y_{2(g)} \rightleftharpoons 2XY_{3(g)}$
$(II)$ $X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)}$
$(III)$ $X_{2(g)} + Z_{2(g)} \rightleftharpoons 2XZ_{(g)}$
$(IV)$ $X_{2(g)} + Y_{4(g)} \rightleftharpoons 2XY_{2(g)}$
$(I)$ $X_{2(g)} + 3Y_{2(g)} \rightleftharpoons 2XY_{3(g)}$
$(II)$ $X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)}$
$(III)$ $X_{2(g)} + Z_{2(g)} \rightleftharpoons 2XZ_{(g)}$
$(IV)$ $X_{2(g)} + Y_{4(g)} \rightleftharpoons 2XY_{2(g)}$
The formation of $NO_2$ in the reaction $2NO + O_2 \rightleftharpoons 2NO_2 + \text{heat}$ is favoured by
Easy
View SolutionFormation of $SO_3$ takes place according to the reaction:
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g); \,\, \Delta H = -45.2 \, kcal$
Which of the following factors favours the formation of $SO_3$?
$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g); \,\, \Delta H = -45.2 \, kcal$
Which of the following factors favours the formation of $SO_3$?
Easy
View SolutionWhich one of the following conditions will favour maximum formation of the product in the reaction
$A_{2(g)} + B_{2(g)} \rightleftharpoons X_{2(g)}; \Delta_{r} H = -X \ kJ$
$A_{2(g)} + B_{2(g)} \rightleftharpoons X_{2(g)}; \Delta_{r} H = -X \ kJ$
MediumNEET 2018
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo