Question

A reaction mixture initially contains $ 2.8M $ $ {H_2}O $ and $ 2.6M $ $ S{O_2} $ . How do you determine the equilibrium concentration of $ {H_2}S $ if $ {K_c} $ for the reaction at this temperature is $ 1.3 \times {10^{ - 6}} $ ?

Answer 1

Hint: Before starting solving the question, we need to write down the equilibrium reaction first and then we will put the values in the formula of equilibrium constant. Equilibrium concentration is basically a concentration where the ratio of the concentration of reactant and the product is constant.

Complete answer:
Now, we know the equilibrium constant for the reaction at a particular temperature is $ 1.3 \times {10^{ - 6}} $ . As we can see that the value of $ {K_c} < 1 $ , this means that most probably the equilibrium mixture contains more reactivity than products.
Let’s assume the concentration of the reactants and products with respects to the time so, we are going to write down the equilibrium reaction, which is:
 $ 2S{O_{2(g)}} + 2{H_2}{O_{(g)}} \rightleftharpoons 2{H_2}{S_{(g)}} + 3{O_{2(g)}} $
   $ 2.6 $ $ 2.8 $ $ 0 $ $ 0 $ $ - - - - $ at $ t = 0 $
 $ 2.6 - 2x $ $ 2.8 - 2x $ $ 2x $ $ 3x $ $ - - - - $ at $ t = t $
Now, we know that equilibrium constant of any reaction can be calculated by dividing the concentration of the products with the concentration of the reactants. Mathematically, it can be written as:
 $ {K_c} = \dfrac{{{{[C]}^c} \times {{[D]}^d}}}{{{{[A]}^a} \times {{[B]}^b}}} $ , where $ {K_c} $ stands for the equilibrium constant, $ A $ and $ B $ stands for the reactants, $ C $ and $ D $ stands for the products formed, $ a,b,c,d $ stands for the number of moles of A,B,C and D and $ [A] $ stands for the equilibrium concentration of A in moles and so on..
Now we will write equilibrium constant for the reaction:
 $ {K_{c = }}\dfrac{{{{[{H_2}S]}^2} \cdot {{[{O_2}]}^3}}}{{{{[S{O_2}]}^2} \cdot {{[{H_2}O]}^2}}} $
After putting the values into the above formula, we get:
 $
  {K_{c = }}\dfrac{{{{[2x]}^2} \cdot {{[3x]}^3}}}{{{{[2.6 - 2x]}^2} \cdot {{[2.8 - 2x]}^2}}} \\
  {K_c} = \dfrac{{4{x^2} \cdot 27{x^3}}}{{{{(2.6 - 2x)}^2} \cdot {{(2.8 - 2x)}^2}}} \\
  {K_c} = \dfrac{{108{x^5}}}{{{{(2.6 - 2x)}^2} \cdot {{(2.8 - 2x)}^2}}} = 1.3 \times {10^{ - 6}} \\
  $
Now, we can see that the value of $ {K_c} $ is very low as compared to the initial concentration of $ {H_2}O $ and $ S{O_2} $ , therefore, $ 2.8 - 2x \approx 2.8 $ and $ 2.6 - 2x \approx 2.6 $ . After rewriting the equation, we get:
 $ 1.3 \times {10^{ - 6}} = \dfrac{{108{x^5}}}{{{{(2.6)}^2} \cdot {{(2.8)}^2}}} $
Now we will solve this equation to find out the value of x and after solving, we get:
 $ x = 0.05767 $
Since, we know that $ 2x $ is the equilibrium concentration for hydrogen sulphide, so we will get:
 $
  [{H_2}S] = 2x = 2 \times 0.05767M = 0.11534M \approx 0.12M \\
  [{H_2}S] = 0.12M \\
  $
So, we will find out the equilibrium concentrations for the two reactants:
 $
  [S{O_2}] = 2.6 - 2x = 2.6 - 2 \cdot 0.05767M = 2.5M \\
  [{H_2}O] = 2.8 - 2x = 2.8 - 2 \cdot 0.05767M = 2.7M \\
  $
As we can see, the concentration of Hydrogen sulphide is very low as compared to the concentration of the reactants.
Therefore, equilibrium concentration of $ [{H_2}S] $ is $ 0.12M $ .

Note:
For solving such types of questions we need to write equilibrium reactions first and then only we will be able to assume their concentration and then the further calculations to find out the result can be done.