Question

Assume that the change in pressure of \({H_2}S\) is small enough to be neglected in the following problem.(a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H₂S with an initial pressure of 0.824 atm.

\(2{H_2}S(g) \rightleftharpoons 2{H_2}(g) + {S_2}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_p} = 2.2 \times {10^{( - 6)}}\)

(b) Show that the change is small enough to be neglected.

Short answer

We can conclude that the change is small enough to be neglected

Step-by-step solution

Define concentration.

It is a constituent divided by total volume of a mixture.

Explanation.

Given:

\(2{H_2}S(g) \rightleftharpoons 2{H_2}(g) + {S_2}(g)\)

We will assume that the change in pressure of is small enough to be neglected.
The equilibrium constant is

a. The equilibrium constant is

Initial pressure of H₂S=0.824 atm

Therefore, pressure regarding H₂ =1.44×10^-2 atm

pressure regarding S₂ =7.2×10^-3 atm

So,

\(\begin{aligned}{{}{}}{{K_p} = \frac{{{{\left( {{P_{{H_2}}}} \right)}^2} \times \left( {{P_{{S_2}}}} \right)}}{{{{\left( {{P_{{H_2}S}}} \right)}^2}}}}\\{ = \frac{{{{\left( {1.44 \times {{10}^{ - 2}}} \right)}^2} \times \left( {7.2 \times {{10}^{ - 3}}} \right)}}{{{{(0.810)}^2}}}}\\{ = 2.27 \times {{10}^{ - 6}}}\end{aligned}\)

Therefore, the equilibrium constant will be 2.27×10^-6

b. Since we get approximately the same value of equilibrium constant, with or without the change of pressure of \({{\rm{H}}_2}{\rm{S}}\), we can conclude that the change is small enough to be neglected.