Question

In: Chemistry

Consider the reaction: CO(g) + H2O(g)<------> CO(g) + H2(g)   and K = 0.118 at 4000 K....

Consider the reaction: CO(g) + H2O(g)<------> CO(g) + H2(g)   and K = 0.118 at 4000 K. A reaction mixture initially contains a CO partial pressure of 1344 mbar and a H2O partial pressure of 1766 mbar at 4000K. Calculate the equilibrium partial pressures of each of the products.

Solutions

Expert Solution

Solution :-

Lets set up the ICE table for the reaction.

CO(g)   +       H2O(g)       <------>     CO2(g)     +        H2(g)   and K = 0.118

1344 mbar 1766mbar                0                    0

-x                   -x                             +x                 +x

1344-x          1766-x                    x                   x

equilibrium constant equation is as follows.

K= [CO2] [H2]/ [CO][H2O]

0.118 = [x][x]/[1344-x][1766-x]

0.118 * [1344-x][1766-x] = x^2

Solving for x we get

X= 393

So the now lets calculate the equilibrium partial pressures of each

[CO] =1344-x = 1344-393 = 951 mbar

[H2O] = 1766 –x = 1766 – 393 = 1373 mbar

[CO2] = x = 393 mbar

[H2] = x = 393 mbar


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