Consider the reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g) Kc=102 at 500 K A reaction mixture initially contains 0.135 M CO...
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Consider the reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g) Kc=102 at 500 KA reaction mixture initially contains 0.135 M CO and 0.135 M H2O.What is the equilibrium concentration of [CO], [H2O], [CO2], and[H2]
Consider the reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g) Kc=102 at 500 KA reaction mixture initially contains 0.135 M CO and 0.135 M H2O.What is the equilibrium concentration of [CO], [H2O], [CO2], and[H2]
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Answer – We are given, reaction - CO(g)+H2O(g) <-----> CO2(g)+H2(g)
Kc = 102 , [CO] = 0.135 M, [H2O] = 0.135 M
We need to put the ICE chart
     CO(g)+H2O(g) <-----> CO2(g)+H2(g)
IÂ Â Â 0.135Â Â Â Â Â 0.135Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â 0
C    -x           -x            +x        +x
E 0.135-x  0.135-x        +x       +x
We know, Kc = [CO2(g)] [H2(g)] / [CO(g)] [H2O(g)]
102 = x *x / (0.135-x) (0.135-x)
So, 102[(0.135-x) (0.135-x)] = x2
102( x2-0.27x+0.0182) =x2
102x2 -27.54x + 1.86 = x2
101x2 -27.54x + 1.86 = 0
Using the quadratic equation
x = 0.123 M
so equilibrium concentration
[CO(g)] =0.135-x
             = 0.135-0.123
             = 0.012 M
[H2O(g)] =0.135-x
             = 0.135-0.123
             = 0.012 M
[CO2(g)] = x = 0.123 M
[H2(g)] = x = 0.123 M
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