Question

At 439 K, this reaction has a Kc value of 0.0569.

2X(g)+2Y(g)−⇀↽−Z(g)

Calculate KpKp at 439 K. Note that the pressure is in units of atmosphere (atm).

Answer 1

summary:

Kp = 0.8581 × 10^ -6

Solution:

We simply used following relation

Kp = Kc × (R ×T )^ n

Where

R = gas constant = 0.0821 L atm/ mole K

T = temperature in Kelvin

n = total mole of gas in product side - total mole of gas in reactant side

For given reaction

n = 1- 4= -3

So we calculate Kp using above equation

Kp= 0.0569 ×( 0.0821 × 493) ^ -3

Kp = 0.8581 × 10^-6