Question

Part A

For the reaction

2A(g)+2B(g)⇌C(g)

Kc = 62.6 at a temperature of 187 ∘C .

Calculate the value of Kp.

Express your answer numerically.

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Part B

For the reaction

X(g)+3Y(g)⇌3Z(g)

Kp = 3.10×10⁻² at a temperature of 325 ∘C .

Calculate the value of Kc.

Express your answer numerically.

Answer 1

relation between Kc and Kp

Kc=Kp / (RT)^Δn

Δn = (Total moles of gas on the products side) - (Total moles of gas on the reactants side)

R = gas constant = 0.0821 L atm/mol-K

T = absolte temperature

use this equation to solve the given questions

Part A

2A(g)+2B(g)⇌C(g)

Δn = 1-4 = -3

Kc = 62.6

T = 273 +187 = 460 K

62.6 = Kp / (0.0821 x 460)^-3

Kp = 62.6 / 53864.54

Kp = 1.16 x 10^-3

PartB

X(g)+3Y(g)⇌3Z(g)

Δn = 3-4 = -1

Kp = 3.10×10−2

T = 273 + 325 = 598K

Kc = 3.10×10⁻² / (0.0821 x 598)^-1

Kc = 1.522