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
CÂ Â Â Â
-x          Â
-x           Â
+x        +x
E 0.135-x Â
0.135-x       Â
+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
            Â
= 0.135-0.123
            Â
= 0.012 M
[H2O(g)] =0.135-x
            Â
= 0.135-0.123
            Â
= 0.012 M
[CO2(g)] = x = 0.123
M
[H2(g)] = x = 0.123
M
