Question

The equilibrium constant for the following reaction is 1.6 ×10⁵ at 1024 K.
\(H_2(g) + Br_2(g) ⇋ 2HBr(g)\)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Answer 1

Given, K_p for the reaction i.e.,
\(H_2(g) + Br_2(g) ↔ 2HBr(g)\) is \(1.6×10^5\)5.
Therefore, for the reaction \(2HBr(g) ↔ H_2(g) + Br(g)\), the equilibrium constant will be,
\(K'p = \frac {1}{K_p}\)
\(K'p= \frac {1}{1.6×10^5}\)
\(K'p= 6.25×10^{-6}\)
Now, let p be the pressure of both H₂ and Br₂ at equilibrium.
                                 \(2HBr(g) ↔ H_2(g) + Br_2(g)\)
Initial conc.                   \(10\)                 \(0\)            \(0\)
At equilibrium          \(10 - 2p\)            \(p\)           \(p\)
Now, we can write,
\(\frac {P_{HBr} × P_2}{P^2_{HBr}} = K'p\)

\(\frac {p × p }{(10 - 2p)^2} = 6.25×10^{- 6 }\)

\(\frac {p}{10 - 2 p} = 2.5 × 10^{- 3}\)
\(P = 2.5 × 10^{-2} - (5.0 × 10^{-3})P\)
\(P + (5.0 × 10^{- 3})P = 2.5 × 10^{- 2}\)
\((1005×10^{- 3}) P = 2.5 × 10^{- 2}\)
\(P = 2.49 × 10^{- 2}\  bar\)
\(P= 2.5×10^{- 2}\  bar\)  (approximately)
Therefore, at equilibrium,
\([H_2] = [Br_2] = 2.49 × 10^{-2}\ bar\)
\([HBr] = 10-2×(2.49×10^{-2})\ bar = 9.95 \ bar = 10\  bar\) (approximately)