The reaction below has an equilibrium constant of Please show me steps, because

Question

The reaction below has an equilibrium constant of

Please show me steps, because I want to understand how to do it.

Explanation / Answer

CO + 2H2 ------> CH3OH .....(1) A) 1/2 CH3OH -> 1/2 CO + H2 .....(2) , we observe that if revere 1 and divide eq by 2 we get eq(2) , hence kp = (1/sqrt kp) = 1/(sqrt(2.26 x10^ 4) = 6.65 x10^ -3 , since Kp = (1/Kp) for reverse reaction and Kp becomes Kp^n ( when reaction is multiplied by n) , based on thses facts Kp for eq(2) is calculated, B) 1/2CO + H2 ---> CH3OH ....(3) , eq(1) multiplied by (1/2) gives eq(3) , hence Kp = Kp^(1/2) = sqrt(KP ) = sqrt(2.26 x10^ 4) = 150.3 , C) 2CH3OH --> 2CO + 4H2 ...(4) , eq(1) reversed and multiplied by 2 we get eq(4) , hence Kp = ( 1/Kp^2) = (1)/(2.26 x10^4)^2 = 1.958 x10^ -9

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