Exam

JEE Main

Year

2011

Subject

Chemistry

Chapter

Chemical Equilibrium

JEE Main 2011 · Chemistry · Chemical Equilibrium

A vessel at 1000 ~K contains CO_2 with a pressure of 0.5 ~atm ....

A vessel at $1000 \mathrm{~K}$ contains $\mathrm{CO}_2$ with a pressure of $0.5 \mathrm{~atm}$. Some of the $\mathrm{CO}_2$ is converted into $\mathrm{CO}$ on the addition of graphite. If the total pressure at equilibrium is $0.8 \mathrm{~atm}$, the value of $\mathrm{K}$ is

Step-by-step solution

Total pressure at equilibrium $=0.8 \mathrm{~atm}$; Total no.of moles $=p+x$. Therefore $\mathrm{p} \propto \mathrm{n} ; \frac{0.5}{0.8}=\frac{\mathrm{p}}{\mathrm{p}+\mathrm{x}} \Rightarrow \mathrm{x}=0.3\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{CO}}^2}{\mathrm{P}_{\mathrm{CO}_2}}=\frac{0.6 \times 0.6}{0.2}=$1.8 $\mathrm{~atm}$