Question

Consider the reaction. 2 upper H upper F (g) double-headed arrow upper H subscript 2 (g) plus upper F subscript 2 (g). At equilibrium at 600 K, the concentrations are as follows. [HF] = 5. 82 x 10-2 M [H2] = 8. 4 x 10-3 M [F2] = 8. 4 x 10-3 M What is the value of Keq for the reaction expressed in scientific notation? 2. 1 x 10-2 2. 1 x 102 1. 2 x 103 1. 2 x 10-3.

Answer 1

The value of the equilibrium constant for the dissociation of HF is $\rm 1.2\;\times\;10^-^3$. Thus, option D is correct.

The given balanced chemical equation is:

$\rm 2\;HF\;\leftrightharpoons H_2\;+\;F_2$

The equilibrium constant for the reaction is given as the ratio of the concentration of product to reactant raised to the stoichiometric coefficient.

Computation of equilibrium constant

The equilibrium constant (Keq) for the given reaction is given as:

$Keq=\rm \dfrac{[H_2]\;[F_2]}{[HF]^2}$

The equilibrium concentration is given as:

• $\rm HF=5.82\;\times\;10^-^2\;M$
• $\rm H_2=8.4\;\times\;10^-^3\;M$
• $\rm F_2=8.4\;\times\;10^-^3\;M$

Substituting the values for the equilibrium constant:

$Keq=\dfrac{[8.4\;\times\;10^-^3]\;[8.4\;\times\;10^-^3]}{[5.82\;\times\;10^-^2]} \\\\Keq=12.1\;\times\;10^-^4\\\\Keq=1.2\;\times\;10^-^3$

The value of the equilibrium constant for the dissociation of HF is $\rm 1.2\;\times\;10^-^3$. Thus, option D is correct.

Learn more about the equilibrium constant, here:

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