The equilibrium constant for the following reaction is $$1.6 \times 10^{5}$$ at 1024 K.

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 1024 K.

Asked by Abhisek | 1 year ago |  73

##### Solution :-

Given, Kp for the reaction i.e., H2 (g) + Br2 (g) ⇒ 2HBr (g) is

$$1.6 \times 10^{5}$$

Therefore, for the reaction   2HBr (g)

⇒ H2 (g) + Br2 (g)  the equilibrium constant will be,

$$K’p= \dfrac{1}{K_{p}}\\ \\ =\dfrac{1}{1.6\times 10^{5}}\\ \\ =6.25\times 10^{-6}$$

Now, let p be the pressure of both H2 and Brat equilibrium.

2HBr (g)     H2 (g) +  Br2 (g)

Initial conc.       10         0       0

At equilibrium   10-2p   p       p

Now, we can write,

$$\dfrac{p_{HBr}\times p_{2}}{p^{2}_{HBr}}=K^{‘}_{p}\\ \\$$

$$\dfrac{p\times p}{(10-2p)^{2}}=6.25\times 10^{-6}\\ \\ \dfrac{p}{10-2p}=2.5\times 10^{-3}\\ \\$$

$$p=2.5\times 10^{-2}-(5.0\times 10^{-3})p\\ \\ p+(5.0\times 10^{-3})$$

$$p=2.5\times 10^{-2}\\ \\ (1005\times 10^{-3})$$

$$=2.5\times 10^{-2}\\ \\$$

$$p=2.49\times 10^{-2}bar$$

$$=2.5\times 10^{-2}bar (approximately)$$

Therefore, at equilibrium,

$$[H_2]=[Br_2]= 2.49\times 10^{-2}bar[HBr]$$

$$= 10-2\times (2.49\times 10^{-2})bar\\ \\$$
$$=9.95 bar=10 bar(approximately)$$

Answered by Pragya Singh | 1 year ago

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