Question

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.

Answer 1

Answer: 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:

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

Initial:                                             7.40

At eqllm:          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