Question

The value of kp for the reaction 2so2 +o2 =2so2 is 5 what will be the partial pressure of o2 at equilibrium when equal moles of so2 and soe are present at equilibrium

Answer 1

Consider the reaction:

2SO₂+O₂----->2SO₂  

Equilibrium constant can be written as:

$K_{P}=\frac{P_{SO_{3}^{2}}}{P_{SO_{2}^{2}}\times P_{O_{2}}}$

As,

$P_{SO_{3}=P_{SO_{2}$

So,

$K_{P}=\frac{1}{P_{O_{2}}}$

$5=\frac{1}{P_{O_{2}}}$

$P_{O_{2}=\frac{1}{5}$

$P_{O_{2}=0.2$