At 423K, this reaction has a Kc value of 0.0659.

Question

At 423K, this reaction has a Kc value of 0.0659. X(g)+Y(g)—><— 2Z(g).

Calculate Kp at 423K. Note that the pressure is units of atmosphere(atm)

Answer 1

K_p = (P_z)² /(P_x)(P_y) and K_c = [Z]²/[X][Y]

Using the ideal gas law, PV = nRT, we can write the following relationship between the pressures used in K_p and the concentrations used in K_c

P = (n/V)RT

Since n/V is just concentration in mol/L we can write

P_x = (n_x/V)RT = [X]RT

Likewise,

P_y = (n_y/V)RT = [Y]RT and P_z = (n_z/V)RT = [Z]RT

So K_p = (P_z)² /(P_x)(P_y) = ([Z]²(RT)²)/ ([X](RT)[Y](RT)) = ([Z]²(RT)²)/ ([X][Y](RT)²)

which means that we can write

K_p = [Z]²/[X][Y] × (RT)²/(RT)² = K_c

Since the (RT) terms cancel out, we have

K_p = K_c

The K_p value that we're asked to calculate is the same as the K_c value we are given.

Answer

At 423K, this reaction has a K_p value of 0.0659.

Note

The (RT) terms don't always cancel. They do in the present case because the number of reactant and product gas molecules are the same. In cases where they differ, K_p and K_c will have different values.

A general formula expressing the relationship between K_p and K_c is

K_p = K_c (RT)^Δn

where Δn = moles of gaseous product – moles of gaseous reactant