Ncert Exercises & Solutions

7.27 The equilibrium constant for the following reaction is 1.6 ×105 at 1024K

H2(g) + Br2(g) 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

NEET SOLUTION:

Given,

K_{p}

for the reaction i.e.,

H_{2 (g)} + {Br}_{2 (g)} ⇌ 2 HBr (g) is 1.6 \times 10^{5} .

Therefore, for the reaction, the equilibrium constant will be,

K'_{p} = \frac{1}{K_{p}} = \frac{1}{1.6 \times 10^{5}} = 6.25 \times 10^{- 6}

Given,

Now, let p be the pressure of both

H_{2} and {Br}_{2}

a equilibrium.

2 {HBr}_{(g)} \leftrightarrow H_{2 (g)} + {Br}_{(g)}

Initial conc . 10 0 0

At equilibrium 10 - 2 p p p

Now, we can write,

\frac{p_{H_{2}} \times p_{{Br}_{2}}}{{p^{2}}_{HBr}} = K'_{p} \frac{p \times p}{{(10 - 2 p)}^{2}} = 6.25 \times 10^{- 6} \frac{p}{10 - 2 p} = 2.5 \times 10^{- 3} p = 2.5 \times 10^{- 2} - (5.0 \times 10^{- 3}) p p + (5.0 \times 10^{- 3}) p = 2.5 \times 10^{- 2} (1005 \times 10^{- 3}) p = 2.5 \times 10^{- 2} p = 2.49 \times 10^{- 2} bar = 2.5 \times 10^{- 2} bar (approx .)

Therefore, at equilibrium,

[H_{2}] = [{Br}_{2}] = 2.49 \times 10^{- 2} bar

[HBr] = 10 - 2 \times (2.49 \times 10^{- 2}) bar

= 9.95 bar = 10 bar (approx .)