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il63 [147K]
4 years ago
7

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2 (g) →2 SO3 (g) At equilibrium, the part

ial pressure of SO2 is 36.9 atm and that of O2 is 16.8 atm. The partial pressure of SO3 is ________ atm.
Chemistry
1 answer:
aniked [119]4 years ago
7 0

Answer : The partial pressure of SO_3 is, 88.84 atm

Solution :  

The given equilibrium reaction is,

2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

\text{Partial pressure of }SO_2 = 36.9 atm

\text{Partial pressure of }O_2 = 16.8 atm

The expression of K_p will be,

K_p=\frac{(p_{SO_3})^2}{(p_{O_2})(p_{SO_2})^2}

Now put all the values in this expression, we get

0.345=\frac{(p_{SO_3})^2}{(16.8)\times (36.9)^2}

(p_{SO_3})=88.84atm

Therefore, the partial pressure of SO_3 is, 88.84 atm

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