Question

On analysis, an equilibrium mixture for the reaction 2h2s(g) 2h2(g) + s2(g) was found to contain 1.0 mol h2s, 4.0 mol h2, and 0.80 mol s2 in a 4.0 l vessel. calculate the equilibrium constant, kc, for this reaction.

Answer 1

Kc = 3.2

The chemical equation is

2H2S ⇌ 2H2 + S2

The equilibrium constant expression is

Kc ={[H2]^2[S2]}/[H2S]^2

1. Calculate the equilibrium concentrations of each component

[H2] = (4.0 mol)/(4.0 L) = 1.0 mol/L

[S2] = (0.80 mol)/(4.0 L) = 0.20 mol/L

[H2S] = (1.0 mol)/(4.0 L) = 0.25 mol/L

2. Calculate the value of the equilibrium constant

Kc = (1.0^2 x 0.20)/0.25^2 = 3.2

Answer 2

The equilibrium constant $\left( {{{\text{K}}_{\text{c}}}}\right)$ for the reaction $2{{\text{H}}_{\text{2}}}{\text{S}}\left(g\right) \rightleftharpoons 2{{\text{H}}_2}\left( g \right) + {{\text{S}}_2}\left( g \right)$ is $\boxed{3.2}$ .

Further Explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

 ${\text{P(g)}} + {\text{Q(g)}}\rightleftharpoons {\text{R(g)}} +{\text{S(g)}}$

Equilibrium constant is the constant that relates to the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for the general reaction is as follows:

${\text{K}}=\dfrac{{\left[ {\text{R}}\right]\left[{\text{S}} \right]}}{{\left[{\text{P}} \right]\left[ {\text{Q}}\right]}}$

Here,

K is the equilibrium constant.

P and Q are the reactants.

R and S are the products.

The given reaction is as follows:

$2{{\text{H}}_{\text{2}}}{\text{S}}\left(g\right)\rightleftharpoons 2{{\text{H}}_2}\left( g \right)+{{\text{S}}_2}\left( g \right)$

The expression for the equilibrium constant for the given reaction is as follows:

${{\text{K}}_{\text{c}}}=\dfrac{{{{\left[ {{{\text{H}}_{\text{2}}}}\right]}^2}\left[ {{{\text{S}}_{\text{2}}}}\right]}}{{{{\left[{{{\text{H}}_2}{\text{S}}}\right]}^2}}}$

                .......(1)

Here,

${{\text{K}}_{\text{c}}}$ is the equilibrium constant.

$\left[{{{\text{H}}_{\text{2}}}}\right]$ is the concentration of hydrogen.

$\left[{{{\text{S}}_{\text{2}}}}\right]$ is the concentration of sulfur.

$\left[{{{\text{H}}_2}{\text{S}}}\right]$ is the concentration of hydrogen sulfide

The equilibrium concentration of ${{\text{H}}_{\text{2}}}$ is calculated as follows:

$\begin{aligned}\left[{{{\text{H}}_{\text{2}}}}\right]&=\dfrac{{{\text{4 mol}}}}{{{\text{4 L}}}}\\&= 1\;{\text{mol/L}}\\\end{aligned}$

The equilibrium concentration of ${{\text{S}}_{\text{2}}}$ is calculated as follows:

$\begin{aligned}\left[{{{\text{S}}_{\text{2}}}} \right]&=\frac{{{\text{0}}{\text{.80 mol}}}}{{{\text{4 L}}}} \\&=0.20\;{\text{mol/L}}\\\end{aligned}$

The equilibrium concentration of ${{\text{H}}_2}{\text{S}}$ is calculated as follows:

$\begin{aligned}\left[{{{\text{H}}_2}{\text{S}}}\right]&=\frac{{{\text{1 mol}}}}{{{\text{4 L}}}}\\&=0.25\;{\text{mol/L}}\\\end{aligned}$

The value of $\left[ {{{\text{H}}_{\text{2}}}}\right]$  is 1 mol/L.

The value of $\left[{{{\text{S}}_{\text{2}}}}\right]$ is 0.20 mol/L.

The value of  $\left[{{{\text{H}}_2}{\text{S}}}\right]$ is 0.25 mol/L.

Substitute these values in equation (1).

$\begin{aligned}{{\text{K}}_{\text{c}}}&=\frac{{{{\left({{\text{1 mol/L}}}\right)}^2}\left( {{\text{0}}{\text{.20 mol/L}}} \right)}}{{{{\left( {{\text{0}}{\text{.25 mol/L}}} \right)}^2}}} \\&=3.2\\\end{aligned}$

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

2. Complete equation for the dissociation of  (aq): brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: H2, S2, H2S, Kc, 3.2, 1 mol, 4 mol, 0.80 moll, 4 L, P, Q, R, S, reactants, products, 1 mol/L, 0.20 mol/L, 0.25 mol/L.