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ad-work [718]
3 years ago
8

In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0

torr, the total pressure at equilibrium is 36.3 torr.
Chemistry
1 answer:
raketka [301]3 years ago
7 0

The question is incomplete , the complete question is ;

In a study of the following reaction :

3Fe(s)+4H_2O(g)\rightleftharpoons Fe_2O_4(s)+4H_2(g)

at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the K_p foe this reaction at 1200 K.

Answer:

The K_p for this reaction at 1200 K is 4.066[/tex].

Explanation:

Partial pressure of the water vapor at equilibrium = p_1=15.0 Torr

Partial pressure of the hydrogen gas at equilibrium  = p_2

Total pressure at equilibrium = P=36.3 Torr

P_1+P_2

p_2=P-P_1=36.3 Torr-15.0 Torr=21.3 Torr

3Fe(s)+4H_2O(g)\rightleftharpoons Fe_2O_4(s)+4H_2(g)

The expression if K_p is given as;

K_p=\frac{(p_2)^4}{(p_1)^4}

(the partial pressure of the gas will be taken along with which partial pressure of the solids are taken as unity)

K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}

K_p=4.066

The K_p for this reaction at 1200 K is 4.066[/tex].

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