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Le Chatelier's Principle for Del N Equals to 0

Consider the ...

Question

Consider the water gas equilibrium reaction,

$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)$

Which of the following statements is true at equilibrium?

If the amount of

C (s)

is increased, less water would be formed

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If the amount of

C (s)

is increased, more

C O

and

H_{2}

would be formed

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If the pressure on the system is increased by halving the volume, more water would be formed

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If the pressure on the system is increased by halving the volume, more

C O

and

H_{2}

would be formed

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Solution

The correct option is C If the pressure on the system is increased by halving the volume, more water would be formed
$K = \frac{[C O (g)] [H_{2} (g)]}{[H_{2} O (g)]}$
Concentration will increase, on halving the volume. There are two terms in numerator. So to keep $K$ constant, concentration of $[H_{2} O]$ should increase much more.

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Similar questions

C O

{C O}_{2}

H_{2} O (g) + C O (g) ⇌ H_{2} (g) + {C O}_{2} (g) + H e a t

C O

H_{2}

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); Δ H = + 130.5 k J

H_{2}

C (s) + H_{2} O ⇌ C O (g) + H_{2} O, Δ H = + 131 k J

H_{2}

2 C_{2} H_{2 (g)} + 5 O_{2 (g)} ⇌ 4 C O_{2 (g)} + 2 H_{2} O_{(g)} △ H = - v e

C O_{2 (g)}

H_{2}

H_{2} (g) + 1 / 2 O_{2} (g) = H_{2} O (g); △ H = - 241.8 k J

C O (g) + 1 / 2 O_{2} (g) = {C O}_{2} (g); △ H = - 283 k J

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