Question

Hydrogen Bromide is formed by a reaction between hydrogen gas and bromine vapor.
Here is the equation, H2(g) + Br2(g) <==> 2HBr(g), with K(eq) = 3.5
Calculate the equilibrium concentration of all gases if 0.40 mol of H2(g) and 0.60 mol of Br2(g) are placed in a 4.0L container.

Answer 1

Question #57687, Chemistry, Other

Hydrogen Bromide is formed by a reaction between hydrogen gas and bromine vapor.

Here is the equation: $\mathsf{H}_{2(\mathsf{g})} + \mathsf{Br}_{2(\mathsf{g})}\leftrightarrow 2\mathsf{HBr}_{(\mathsf{g})}$ , with $\mathsf{K}_{(\mathsf{eq})} = 3.5$

Calculate the equilibrium concentration of all gases if 0.40 mol of $\mathsf{H}_{2(\mathsf{g})}$ and 0.60 mol of $\mathsf{Br}_{2(\mathsf{g})}$ are placed in a 4.0 L container.

Answer:

$$K _ {e q} = \frac {[ H B r ] ^ {2}}{[ H _ {2} ] [ B r _ {2} ]}$$

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

0.60 mol $\mathrm{Br}_2 / 4.0\mathrm{L} = 0.15\mathrm{M}\mathrm{Br}_2$

$3.5 = [(2x)^{2}] / [(0.15 - x)(0.10 - x)]$

$3.5 = (4x^{2}) / (0.015 - 0.15x - 0.10x + x^{2})$

$3.5 = (4x^{2}) / (0.015 - 0.25x + x^{2})$

0.0525 - 0.875x + 3.5x² = (4x²)

0.0525 - 0.875x - 0.5x² = 0

$-0.5x^{2} - 0.875x + 0.0525 = 0$

X = 0.06

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