Question

Please Answer Both Parts.

Part A

For the reaction

3A(g)+3B(g)⇌C(g)

Kc = 44.0 at a temperature of 93 ∘C .

Calculate the value of Kp.

Express your answer numerically.

Part B

For the reaction

X(g)+3Y(g)⇌3Z(g)

Kp = 3.20×10⁻² at a temperature of 351 ∘C .

Calculate the value of Kc.

Express your answer numerically.

Answer 1

A)
T= 93.0 oC
= (93.0+273) K
= 366 K
Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant
Δ n = -5

use:
Kp= Kc (RT)^Δ n
Kp = 44*(0.08206*366.0)^(-5)
Kp = 1.80*10^-6
Answer: 1.80*10^-6

B)
T= 351.0 oC
= (351.0+273) K
= 624 K
Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant
Δ n = -1

use:
Kp= Kc (RT)^Δn
3.2*10^-2 = Kc *(0.08206*624.0)^(-1)
Kc = 1.639
Answer: 1.64