Given:

For the given reaction,

H₂(g) + Br₂(g) ⇌ 2HBr(g)

The equilibrium constant(K_c) is 1.6 × 10⁵

Initial pressure of HBr = 10bar

Temperature = 1024K

For equilibrium constant in terms of pressure, we apply the formula given below:

K_p =

Where P_x and P_Y are the partial pressures of x and y(products)

P_A and P_B are the partial pressures of a and b (reactants)

As we know that the relationship between K_p and K_c is K_p=K_c (R/T)^Δn

Where Δn = no. of moles of products – no. of moles of reactants

Hence for the given reaction,

Δn = 0

∴ K_p = K_c

As the given reaction is a reversible reaction. Hence, we can write

2HBr(g) ⇌ H₂(g) + Br₂(g), ∴ K_c =

Initial 10bar

At

Equilibrium 10-P P/2 P/2

Using the above formula,

K_p =

K_p = = K_c

⇒ =

⇒ =

Taking square root of both sides, we get

⇒ =

⇒4.0 × 10² = 2(10-P)

⇒402 P = 20

⇒P =

⇒P = 4.97 × 10^-2 bar

Hence, at equilibrium

P_H2 = P_Br2 = P/2

⇒ = 2.48 × 10^-2 bar

P_HBr = 10-P ≈ 10 bar

Thus, the equilibrium pressure of H₂ and Br₂ is 2.48 × 10^-2 bar and HBr is 10 bar