For the reaction $Mg(HCO_3)_{2(s)} \rightleftharpoons MgCO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$,the equilibrium constant $K_p = 64 \ atm^2$. Calculate the total pressure at equilibrium.
(16 ATM) The equilibrium expression for the reaction is $K_p = P_{CO_2} \times P_{H_2O}$.
Since the stoichiometry of the gaseous products $CO_2$ and $H_2O$ is $1:1$,their partial pressures at equilibrium will be equal,i.e.,$P_{CO_2} = P_{H_2O} = P$.
Thus,$K_p = P \times P = P^2$.
Given $K_p = 64 \ atm^2$,we have $P^2 = 64$,which gives $P = 8 \ atm$.
The total pressure at equilibrium is $P_{total} = P_{CO_2} + P_{H_2O} = 8 \ atm + 8 \ atm = 16 \ atm$.
Since the stoichiometry of the gaseous products $CO_2$ and $H_2O$ is $1:1$,their partial pressures at equilibrium will be equal,i.e.,$P_{CO_2} = P_{H_2O} = P$.
Thus,$K_p = P \times P = P^2$.
Given $K_p = 64 \ atm^2$,we have $P^2 = 64$,which gives $P = 8 \ atm$.
The total pressure at equilibrium is $P_{total} = P_{CO_2} + P_{H_2O} = 8 \ atm + 8 \ atm = 16 \ atm$.
Explore More
Similar Questions
$\Delta n$,the change in the number of moles for the reaction,$C_{12}H_{22}O_{11(s)} + 12O_{2(g)} \rightleftharpoons 12CO_{2(g)} + 11H_2O_{(l)}$ at $25 \ ^\circ C$ is
Easy
View SolutionIn a $500 \,mL$ flask,the degree of dissociation of $PCl_5$ at equilibrium is $40 \%$ and the initial amount is $5 \,moles$. The value of equilibrium constant in $mol \,L^{-1}$ for the decomposition of $PCl_5$ is
DifficultAP EAMCET 2008
View SolutionFor the dissociation reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the degree of dissociation $(\alpha)$ in terms of $K_p$ and total equilibrium pressure $P$ is
Medium
View SolutionFor the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the value of $K_p$ at $800 \ ^oC$ is $1.16 \ atm$. If $1 \ mol$ of $CaCO_{3(s)}$ is taken in a $1 \ L$ container to start the reaction,what will be the partial pressure of $CO_2$ at equilibrium in $atm$?
Medium
View SolutionAt $300 \ K$,$K_C$ for the reaction $A_2B_{2(g)} \rightleftharpoons A_{2(g)} + B_{2(g)}$ is $100 \ mol \ L^{-1}$. What is its $K_p$ (in $atm$) at the same temperature? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
Vedclass 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