Question
The equilibrium constant for the reaction 2SO2 + O2 ⇋ 2 SO3 is 900 atm−1 at 800K. A mixture containing SO3 and O2 having initial pressures of 1atm and 2atm is heated to equilibrate. The partial pressure of O2at equilibrium will be
Answer: Option C
:
C
Correct option is c). Considering the backward reaction we have,
2SO3⇋2SO2+O2KP=1/900atmInitialpressure1atm02atmPressureatequilibrium1−xx2+x/2
Let partial pressure of SO2=p1and partial pressure of O2=p2and partial pressure of SO3=p3
K<p=(p1)2×(p2)/(p3)2=x2(2+x/2)/(1−x)2=1/900
AsKpfor this reaction is very small, x<<1.
Taking 2+x/2 as x and (1−x) as x and solving for x, we get
X=0.0236
Hence, at equilibrium, partial pressure of SO3=1−X=0.9764atm,
Partial pressure of SO2=x=0.0236atm
And Partial pressure of O2=2+x/2=2.0118atm
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:
C
Correct option is c). Considering the backward reaction we have,
2SO3⇋2SO2+O2KP=1/900atmInitialpressure1atm02atmPressureatequilibrium1−xx2+x/2
Let partial pressure of SO2=p1and partial pressure of O2=p2and partial pressure of SO3=p3
K<p=(p1)2×(p2)/(p3)2=x2(2+x/2)/(1−x)2=1/900
AsKpfor this reaction is very small, x<<1.
Taking 2+x/2 as x and (1−x) as x and solving for x, we get
X=0.0236
Hence, at equilibrium, partial pressure of SO3=1−X=0.9764atm,
Partial pressure of SO2=x=0.0236atm
And Partial pressure of O2=2+x/2=2.0118atm
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