Chemistry
posted by Julia .
Calculate the value of Kp for the equation
C(s)+CO2(g)<>2CO(g) Kp=?
given that at a certain temperature
C(s)+2H2O(g)<>CO2(g)+2H2(g) Kp1=3.33
H2(g)+CO2(g)<>H2O(g)+CO(g) Kp2=.733
Please show detailed steps.

C(s)+2H2O(g)<>CO2(g)+2H2(g) Kp1=3.33
H2(g)+CO2(g)<>H2O(g)+CO(g) Kp2=.733
Multiply equation 2 by 2 and add to equation 1 to obtain:
C(s) + 2H2O + 2H2 + 2CO2 ==> CO2 + 2H2 + 2H2O + 2CO
Notice 2H2O cancels, 2H2 cancels, 1 CO2 cancels to leave
C(s) + CO2 ==> 2CO
Kp for eqn 1 is Kp1 as listed. Kp for equation 2 multiplied by 2 is K^2p and is K'p2 (0.733)^2. That's because when you multiply and equation by a coefficient(n) the k values is k^n.
When we ADD equations (as we did above), the combined k value is multiplied; i.e., kp1 x (Kp2)^2 = ? 
Thank you so much for the detailed steps. Finally understood what I need to do.