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VITEEE 2011 Paper

Section: Chemistry

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Question : 73 of 120

Marks: +1, -0

A vessel at 1000 K contains

\mathrm{CO}_2

with a pressure of 0.5 atm. Some of the

\mathrm{CO}_2

is converted into CO on the addition of graphite. The value of K if the total pressure at equilibrium is 0.8 atm, is

1.8 atm
3 atm
0.3 atm
0.18 atm

Solution:

Initial Equilibrium =

\underset{0.5}{\mathrm{CO}_2 + c}_{0.5 - x} \rightleftharpoons \underset{0}{2\mathrm{CO}}_{2x}

Total pressure at Equilibrium

=0.8

atm

Now

(0.5-x)+2x=0.8

atm

\therefore \;\; x=0.8-0.5=0.3

Pressure of

C \mathrm{O}_2

at equilibrium

=0.5-0.3

=0.2

atm

Pressure of

C O

at equilibrium

=2 \times 0.3=0.6

atm

\therefore K=\frac{(pco)^2}{pco_2}=\frac{(0.6)^2}{0.2}=\frac{0.36}{0.2}=1.8\ \text{atm}

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