Question

Good day,

For the reaction
3A(g)+2B(g)⇌C(g)
Kc= 72.2 at a temperature of 237 ∘C
Calculate the value of Kp
Express your answer numerically.

Thanks

Answer 1

K_p = K_c×(R×T)^Δn

There is 1 mol of gas on the product side of the chemical equation, and there are 3+2=5 mol of gaseous reactants, so Δn = (1-5) = -4 in this case.

T = 237 + 273.15 = 510.15 K

R = 0.08206 L×atm/mol×K

K_p = K_c×(R×T)^Δn = 72.2×(0.08206×510.15)^-4 = 72.2×(41.86)^-4 = 72.2×3.25×10^-7 = 234.65×10^-7 = 2.34×10^-5

Answer: 2.34×10^-5