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liraira [26]
3 years ago
10

The equilibrium constant for the following reaction: H2(g) + Br2(g) ↔ 2HBr (g) is 1.10 x 103 at a certain temperature. Find the

equilibrium pressure of HBr if 7.40 atm of HBr is introduced into a sealed container at this temperature.
Chemistry
1 answer:
mihalych1998 [28]3 years ago
7 0

<u>Answer:</u> The partial pressure of HBr at equilibrium is 6.98 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of HBr = 7.40 atm

As, initially HBr is present. So, the reaction is proceeding backwards.

For the given chemical equation:

                     H_2(g)+Br_2(g)\rightarrow 2HBr(g)

<u>Initial:</u>                                             7.40

<u>At eqllm:</u>          x            x          7.40-2x

The expression of K_p for above equation follows:

K_p=\frac{(p_{HBr})^2}{p_{H_2}\times p_{Br_2}}

We are given:

K_p=1.10\times 10^3

Putting values in above expression, we get:

1.10\times 10^3=\frac{(7.40-2x)^2}{x\times x}\\\\x=-0.237,0.210

Neglecting the negative value of 'x' because partial pressure cannot be negative.

So, equilibrium partial pressure of HBr = (7.40 - 2x) = [7.40 - 2(0.210)] = 6.98 atm

Hence, the partial pressure of HBr at equilibrium is 6.98 atm

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