Question

The reaction below has an equilibrium constant of

Kp=2.26×104 at 298 K.
CO(g)+2H2(g)⇌CH3OH(g)
Part A: Calculate Kp for the reaction below.
1/2CH3OH(g)⇌1/2CO(g)+H2(g)

Answer 1

Answer: 6.65*10^-3

Step-by-step explanation:

The reaction below has the products and reactants reversed, so the Kp will be inversed (Kp^-1). The coefficients are also halved, so the Kp^-1 will be to the power of 1/2. This means that the Kp for the reaction below is
`(K_p^(-1))^(1/2)` =
`K_p^{-(1)/(2) }` =
`(1)/(√(K_p))` =
`(1)/(√(2.26*10^4))` = 6.65*10^-3