Chemistry
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AiH,,AiH, and AiH, of an element M are 750, 1500 and 7750 kJ mol respectively. The

AiH,,AiH, and AiH, of an element 'M' are 750, 1500 and 7750 kJ mol respectively. The stable formula
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leeleegavin212
leeleegavin212
4,4(27 marks)

This reaction has an equilibrium constant of K_p=2.26\times 10^4 at 298K.

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

Calculate Kp for each reaction and predict whether reactant or products will be favoured at equilibrium.

A) CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)

B) \frac{1}{2}CO(g)+H_2(g)\rightleftharpoons \frac{1}{2}CH_3OH(g)

C) 2CH_3OH(g)\rightleftharpoons 2CO(g)+4H_2(g)

A) CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g) : 0.442\times 10^-4} : reactants are favoured

B) \frac{1}{2}CO(g)+H_2(g)\rightleftharpoons \frac{1}{2}CH_3OH(g): 1.50\times 10^2} : products are favoured

C) 2CH_3OH(g)\rightleftharpoons 2CO(g)+4H_2(g) : 4.42\times 10^{-9}: reactants are favoured

Explanation:

1. We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)

K_{p}'=(\frac{1}{K_p})

K_{p}'=(\frac{1}{2.26\times 10^4})=0.442\times 10^-4}

As K_p' is less than 1, reactants will be favored at equilibrium.

2. We need to calculate the equilibrium constant for the half equation of above chemical equation, which is:

\frac{1}{2}CO(g)+H_2(g)\rightleftharpoons \frac{1}{2}CH_3OH(g)

K_{p}'=(\sqrt{K_p})

K_{p}'=(\sqrt{2.26\times 10^4})=1.50\times 10^2}

As K_p' is greater than 1, products will be favored at equilibrium.

3.  We need to calculate the equilibrium constant for the reverse and twice the equation of above chemical equation, which is

2CH_3OH(g)\rightleftharpoons 2CO(g)+4H_2(g)

K_{p}'=(\frac{1}{K_p})^2

K_{p}'=(\frac{1}{2.26\times 10^4})^2=4.42\times 10^{-9}

As K_p' is less than 1, reactants will be favored at equilibrium.

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