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.

The partial pressure of HBr at equilibrium is 6.98 atm

Explanation:

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:

Initial:                                             7.40

At eqllm:          x            x          7.40-2x

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

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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