Question

At equilibrium for the reaction H2 (g) + Br2 (g) is revesible arrow 2HBr (g) in a 10 litre vessel was found to contain 2.5 X 10 to the power -3 mole of H2 , 0.150 mole of HBr and 2.8 x 10 to the power-3 mole of BR2. What is the value of k at this temperature?

Answer 1

Question #48734, Chemistry, Other

At equilibrium for the reaction H2 (g) + Br2 (g) is reversible arrow 2HBr (g) in a 10 litre vessel was found to contain 2.5 X 10 to the power -3 mole of H2, 0.150 mole of HBr and 2.8 x 10 to the power-3 mole of BR2. What is the value of k at this temperature?

Answer:

$$\mathrm{H_2} + \mathrm{Br_2} = 2\mathrm{HBr}$$ $$V = 10L$$ $$n(\mathrm{H_2}) = 2.5 \cdot 10^{-2} \text{ mol}$$ $$n(\mathrm{HBr}) = 0.150 \text{ mol}$$ $$n(\mathrm{Br_2}) = 2.8 \cdot 10^{-2} \text{ mol}$$

Because the equilibrium constant is the ratio and the reaction takes place in a constant volume, the concentration can be used instead of the number of particles in order to find the constants:

$$K = \frac{HBr)^2}{2[Br_2]} = 0.15^2 / (2.5 \cdot 10^{-2} \cdot 2.8 \cdot 10^{-2}) = 32.14$$

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