Question

Hydrogen bromide is formed by a reaction between hydrogen gas and bromine vapor in accordance with the following equation: H2 + Br2 = 2HBr
Calculate the equilibrium concentrations of all the gases if 0.40 mole H2 and 0.60 mole of Br2 are placed in a container 4,00 L.

Answer 1

Question #60234, Chemistry, Other

Hydrogen bromide is formed by a reaction between Hydrogen gas and Bromine vapor in accordance with the following equation: $H_{2} + Br_{2} = 2HBr$

Calculate the equilibrium concentrations of all the gases if 0.40 mole $H_{2}$ and 0.60 mole of $Br_{2}$ are placed in a container 4,00 L.

Answer:

$I =$ initial concentration; $C =$ change; $E =$ equilibrium

$K_{(eq)}$ for this reaction is 3.5.

$0.40\ \mathrm{mol}\ H_{2} / 4.0\ \mathrm{L} = 0.10\ \mathrm{M}\ H_{2}$

$0.60\ \mathrm{mol}\ Br_{2} / 4.0\ \mathrm{L} = 0.15\ \mathrm{M}\ Br_{2}$

We don't know the change, so we put $x^*$ the mole ratio.

$$\ldots H_{2}(g) + Br_{2}(g) \leftrightarrow 2HBr(g)$$

I...0.10 M...0.15 M...0 M

C...-x...-x...+2x

E...0.10-x...0.15-x...2x

Now since 3.5 is not puny, the $x$'s cannot be ignored. If K were to be, for example, $1 \times 10^{-5}$, then sure, ignore the $x$'s. But it's not so you can't.

$$K_{(eq)} = \frac{[HBr]^{2}}{[H_{2}][Br_{2}]}$$ $$3.5 = \frac{[2x]^{2}}{(0.15-x)(0.10-x)}$$ $$3.5 = \frac{4x^{2}}{(0.015-0.15x-0.10x+x^{2})}$$ $$3.5 = \frac{4x^{2}}{(0.015-0.25x+x^{2})}$$ $$0.0525 - 0.875x + 3.5x^{2} = (4x^{2})$$ $$0.0525 - 0.875x - 0.5x^{2} = 0$$ $$-0.5x^{2} - 0.875x + 0.0525 = 0$$

After solving the quadratic equation, we receive:

$$x_{1} = -1.81; \ x_{2} = 0.06$$

-1.81 is negative, so it will not be taken into further calculations.

$$2 \cdot 0.06 = 0.12 = 0.12\ \mathrm{M}\ \mathrm{HBr}$$ $$0.10 - 0.06 = 0.04 = 0.04\ \mathrm{M}\ \mathrm{H}_{2}$$ $$0.15 - 0.06 = 0.09 = 0.09\ \mathrm{M}\ \mathrm{Br}_{2}$$

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