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Standard XII

Chemistry

Characteristics of Equilibrium Constant

A vessel; at ...

Question

A vessel; at 1000 K contains $C O_{2}$ with a pressure of 0.5 atm. Some of the $C O_{2}$ is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is:

1.8 atm

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0.18 atm

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Solution

The correct option is A 1.8 atm
$C O_{2} + C_{g r a p h i t e} ⇌ 2 C O$
$P_{i n i t i a l} 0.5 a t m 0$
$P_{f i n a l} (0.5 - x) a t m 2 x a t m$
$Total P at equilibrium 0.5 - x + 2 x = 0.5 + x a t m$
$0.8 = 0.5 + x$
$∴ x = 0.8 - 0.5 = 0.3 a t m$
Now $K_{p} = \frac{(P_{C O})^{2}}{P_{C O_{2}}}$
$K_{p} = \frac{(2 \times 0.3)^{2}}{(0.5 - 0.3)} = \frac{(0.6)^{2}}{(0.2)} = 1.8 a t m$

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Characteristics of Equilibrium Constant

Standard XII Chemistry

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A vessel; at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is:

The correct option is A 1.8 atm CO2+Cgraphite⇌2CO Pinitial 0.5 atm 0 Pfinal (0.5−x)atm 2x atm Total P at equilibrium 0.5−x+2x=0.5+x atm 0.8=0.5+x ∴x=0.8−0.5=0.3 atm Now Kp=(PCO)2PCO2 Kp=(2×0.3)2(0.5−0.3)=(0.6)2(0.2)=1.8 atm

A vessel; at 1000 K contains $C O_{2}$ with a pressure of 0.5 atm. Some of the $C O_{2}$ is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is: