Question

For the reaction 2NO(g) + O₂(g) ⇄ 2NO₂(g) ∆_rxnG⁰ = −72.6 kJ/mol At 600 K, if a reaction vessel initial has partial pressures of p_NO2(g) = 107 Pa, p_NO(g) = 2 Pa, and pO2(g) = 2Pa, in which direction will the reaction occur, forwards or backwards?

Answer 1

step 1: find Kc

T = 600 K

G = -72.6 KJ/mol

G = -72600 J/mol

use:

ΔG = -R*T*ln Kc

-72600 = - 8.314*600.0* ln(Kc)

ln Kc = 14.5538

Kc = 2.092*10^6

step 2: find Kp

T = 600 K

Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant

Δ n = -1

Kp= Kc (RT)^Δ n

Kp = 2092000.0*(8.314*600.0)^(-1)

Kp = 4.194*10^2

Step 3: find Qp and predict

Qp = p(NO2)^2 / p(NO)^2*p(O2)

Qp = 107^2 / (2^2 * 2)

Qp = 107^2 / 8

Qp = 1.43*10^3

Since Qp is greater than Kp, the reaction will occur in backward direction

Answer: abckward