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Question: The equilibrium constant, Kp for the reaction 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g) is 4....

The equilibrium constant, Kp for the reaction

2SO2(g) + O2(g) ⇌\rightleftharpoons 2SO3(g)

is 4.0 atm–1 at 1000 K. What would be the partial pressure of O2 if at equilibrium the amount of SO2 and SO3 is the same ?

A

16.0 atm

B

0.25 atm

C

1 atm

D

0.75 atm

Answer

0.25 atm

Explanation

Solution

2SO2 (g) + O2 ⇌\rightleftharpoons 2SO3(g)

Kp = 4.0 atm–2

Kp=(SO3)3(SO2)2(O2)K_{p} = \frac{\left( SO_{3} \right)^{3}}{\left( SO_{2} \right)^{2}\left( O^{2} \right)}

Given that at equilibrium the amount of SO2 and SO3 is the same so

(SO3)2(SO2)2(O2)=4\frac{\left( SO_{3} \right)^{2}}{\left( SO_{2} \right)^{2}\left( O^{2} \right)} = 4

⇒\Rightarrow [O2]=\left[ \mathrm { O } _ { 2 } \right] = 14\frac{1}{4} = 0.25 atm.