# The value of Kc for rxn below 2 H2S(g) <----> 2H2(g) + S2 (g) find value of [H2S] ?

## The value of Kc for rxn below 2 H2S(g) <----> 2H2(g) + S2 (g) is 2.2 x 10^-4 at 1400 K. A sample of [H2S] = 5.00 M is heated to 1400 K in a sealed vessel. After chemical equilibrium, find value of [H2S] No H2 or S2 was present original sample.

Apr 11, 2018

The equilibrium concentration of ${H}_{2} S$ is $4.78 M$

#### Explanation:

${K}_{c} = 2.2 \times {10}^{-} 4 = \frac{{\left[{H}_{2}\right]}^{2} \left[{S}_{2}\right]}{{\left[{H}_{2} S\right]}^{2}}$

Let

${n}_{{H}_{2} S}^{0}$ = the initial concentration of ${H}_{2} S = 5 M$

Assume that $x$ molar concentration of ${H}_{2} S$ decomposes at equilibrium. At equilibrium, the concentration of ${H}_{2} S$ will be ${n}_{{H}_{2} S}^{0} - x$.

This will produce $x$ molar concentration of ${H}_{2}$ and $\frac{x}{2}$ molar concentration of ${S}_{2}$. Thus at equilibrium we will have

${K}_{c} = \frac{{x}^{2} \left(\frac{x}{2}\right)}{{\left({n}_{{H}_{2} S}^{0} - x\right)}^{2}}$

Because the equilibrium constant is small, we will make the approximation that ${n}_{{H}_{2} S}^{0} - x \approx {n}_{{H}_{2} S}^{0}$. This allows us to write

${K}_{c} \approx \frac{{x}^{3}}{2 {\left({n}_{{H}_{2} S}^{0}\right)}^{2}}$

Solving for $x$ we have

$x = {\left[2 {K}_{c} {\left({n}_{{H}_{2} S}^{0}\right)}^{2}\right]}^{\frac{1}{3}} = {\left[2 \cdot 2.2 \times {10}^{-} 4 {\left(5\right)}^{2}\right]}^{\frac{1}{3}} \approx 0.22$

This would give the final concentration of ${H}_{2} S$ as

${n}_{{H}_{2} S}^{0} - x = 5 - 0.22 = 4.78 M$

As a check of our approximation we can calculate the equilibrium constant as

$\frac{{0.22}^{2} \left(\frac{0.22}{2}\right)}{{\left(4.78\right)}^{2}} \approx 2.3 \times {10}^{-} 4$

which is less than a 5% error from the given value.