Question

The equilibrium constant, Kp for the reaction 2SO2(g) + O2(g) = 2S03(g) is 4.0 atm-' at 1000 K. What
would be the partial pressure of O2 if at equilibrium the amount of SO2 and SO3 is the same?
(A) 16.0atm
(B) 0.25 atm
(C) 1 atm
(D) 0.75 atm​

Answer 1

Answer:

0.25 atm

Explanation:

Here Pso3=Pso2

We know that Kp=(Pso3)^2/(Pso2)^2*Po2

4=1/Po2

ThereforePo2=1/4=0.25

Answer 2

The partial pressure of oxygen at equilibrium will be 0.25 atm

Explanation:

For the given chemical equation:

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

The expression of $K_p$ for the above equation follows:

$K_p=\frac{p_{SO_3}^2}{p_{SO_2}^2\times p_{O_2}}$

We are given:

$K_p=4.0\\\\p_{SO_3}=p_{SO_2}$

Now, the expression of equilibrium constant becomes:

$K_p=\frac{1}{p_{O_2}}$

Putting values in above equation, we get:

$4=\frac{1}{p_{O_2}}\\\\p_{O_2}=\frac{1}{4}=0.25$

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