Question

At a particular temperature, K = 1.00×102 for the reaction: H2(g) + F2(g)= 2HF(g) In an experiment, at this temperature, 1.40 mol of H2 and 1.40 mol of F2 are introduced into a 1.14-L flask and allowed to react. At equilibrium, all species remain in the gas phase. What is the equilibrium concentration (in mol/L) of H2?

Answer 1

Answer:

0.20175 mol/L

Explanation:

The given equilibrium reaction and the equilibrium concentrations are shown below as:-

$\begin{matrix}&H_2&+&F_2&\rightleftharpoons &2HF\\ At\ time, t = 0 &1.40&&1.40&&0\\At\ time, t=t_{eq}&-x&&-x&&+2x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.40-x&&1.40-x&&2x\end{matrix}$

The Kc of an equilibrium reaction measures relative amounts of the products and the reactants present during the equilibrium.

It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.

The expression for the Kc is:-

$K_c=\frac{[HF]^2}{[H_2][F_2]}$

Given that:- Kc = $1.00\times 10^2$

Thus, applying the values as:-

$1.00\times 10^2=\frac{(2x)^2}{(1.40-x)(1.40-x)}$

$10^2=\frac{4x^2}{\left(1.4-x\right)^2}$

$100\left(1.4-x\right)^2=4x^2$

Solving, we get that:-

$x=1.17$

Thus,  

$[H_2]_{eq}=1.40-x=1.40-1.17=0.23\ moles$

Given that:- Volume = 1.14 L

So, $Molarity=\frac{0.23}{1.14}\ mol/L=0.20175\ mol/L$