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Atmospheric Pressure

For the equil...

Question

For the equillibrium $C (s) + C O_{2} (g) ⇋ 2 C O (g)$
$K_{p} = 63 a t m$ $a t 1000 K$ .
If at equillibrium $P_{C O} = 10 P_{C O_{2}}$ then total pressure at equillibrium is :

6.30 atm

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0.693atm

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6.93 atm

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69.3atm

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Solution

The correct option is C 6.93 atm
Let $P_{C O 2} = x a t m$
$P_{C O} = 10 P_{C O 2} = 10 x a t m$
The equilibrium constant $K_{p} = \frac{P_{C O}^{2}}{P_{C O 2}}$
$63 = \frac{(10 x)^{2}}{x}$
$63 x = 100 x^{2}$
$100 x^{2} - 63 x = 0$
This is quadratic equation with solution $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ 63
$x = \frac{- (- 63) \pm \sqrt{(- 63)^{2} - 4 (100) (0)}}{2 (100)}$
$x = \frac{63 \pm 63}{200}$
$x = 0$ or $x = 0.63$
But partial pressure cannot be 0.
Hence, $P_{C O 2} = x = 0.63 a t m$
$P_{C O} = 10 P_{C O 2} = 10 x = 10 \times 0.63 = 6.3 a t m$
Total pressure $= 6.3 + 0.63 = 6.93 a t m$
Note: Since carbon is in solid state, its pressure is not included in equilibrium constant expression.

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