Question

What is the Kp of this reaction at 700oC if the Kc is 0.40? 

CO(g) + 2H2(g)  CH3OH(g)

A)5.34 x 10-5

B) 6.27 x 10-5

C) 8.98 x 10-5

D) 1.76 x 10-5

E) 9.66 x 10-5

Answer 1

i am pretty sure its D

Answer 2

The answer is B) 6.27 x 10⁻⁵

Step-by-step explanation:

The relation between Kp and Kc is given by the following equation:

Kp= Kc (RT)ⁿ

We have:

Kc= 0.40

T= 700ºC= 973K

R=0,082 L.atm/K.mol

To obtain n (or usually called Δn which is the change in mol), we have to look at the number of mol of products (CH₃OH) and reactants (CO and H₂) in gaseous phase.

n= Δn= (total number of mol of products) - (total number of mol of reactants)

n=Δn= (mol CH₃OH) - (mol CO + mol H₂)= 1 - (1+2)= 1-3= -2

So, we introduce the values in the equation to obtain Kp:

Kp= Kc (RT)ⁿ

Kp= 0.40 (0.082 x 973)⁻²

Kp= 6.27 x 10⁻⁵