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Question: At a given temperature, \({{K}_{c}}\) is 4 for the reaction: \[{{H}_{2}}(g)+C{{O}_{2}}(g)\Leftrigh...

At a given temperature, Kc{{K}_{c}} is 4 for the reaction:
H2(g)+CO2(g)⇔H2O(g)+CO(g){{H}_{2}}(g)+C{{O}_{2}}(g)\Leftrightarrow {{H}_{2}}O(g)+CO(g)
Initially, 0.6 moles each of H2{{H}_{2}} and CO2C{{O}_{2}} are taken in a 1-liter flask. The equilibrium concentration of H2O(g){{H}_{2}}{{O}_{(g)}} is:

Explanation

Solution

We know that when the rate of forward reaction in a system is equal to the rate of backward reaction, the system is said to be in equilibrium. The concentration of the substances involved in the reaction at equilibrium is known as the equilibrium concentration of that substance.

Complete step-by-step answer: For the given reaction, it is given to us that the equilibrium constant Kc{{K}_{c}} for the reaction is 4. So,
Kc=[CO][H2O][H2][CO2]=4{{K}_{c}}=\dfrac{[CO][{{H}_{2}}O]}{[{{H}_{2}}][C{{O}_{2}}]}=4
Now, initially, the concentration of reactants H2{{H}_{2}} and CO2C{{O}_{2}} is 0.6 moles each. The concentration of the products H2O{{H}_{2}}O and CO is 0.
Now, let us assume that x moles of each reactant are utilized to form products when the reaction reaches equilibrium.
So, at equilibrium, the concentration of reactants H2{{H}_{2}} and CO2C{{O}_{2}} is (0.6-x) moles each. The concentration of the products H2O{{H}_{2}}O and CO is x moles each.
So, we can say that
Kc=(x)(x)(0.6−x)(0.6−x)=4{{K}_{c}}=\dfrac{(x)(x)}{(0.6-x)(0.6-x)}=4
Upon solving this equation, we get

x2=4×(0.6−x)2 x2=4×(x2−1.2x+0.36) x2=4x2−4.8x+1.44  {{x}^{2}}=4\times {{(0.6-x)}^{2}} \\\ {{x}^{2}}=4\times ({{x}^{2}}-1.2x+0.36) \\\ {{x}^{2}}=4{{x}^{2}}-4.8x+1.44 \\\

The equation needs to be factorized to find the values of x.

3x2−4.8x+1.44=0 3(x2−1.6x+0.48)=0 3(x2−0.4x−1.2x+0.48)=0 3(x(x−0.4)−1.2(x−0.4))=0 3(x−1.2)(x−0.4)=0  3{{x}^{2}}-4.8x+1.44=0 \\\ 3({{x}^{2}}-1.6x+0.48)=0 \\\ 3({{x}^{2}}-0.4x-1.2x+0.48)=0 \\\ 3(x(x-0.4)-1.2(x-0.4))=0 \\\ 3(x-1.2)(x-0.4)=0 \\\

So, the values of x are 1.2 and 0.4
Since the initial moles taken are 0.6, x is 0.4. The value of x cannot be 1.2 as the moles of products formed cannot be more than the moles of reactants taken.
We know that the flask contains 1 liter of solution, so the equilibrium concentration of H2O{{H}_{2}}O can be calculated by
[H2O]=nV=x1L=0.4mol1L=0.4M[{{H}_{2}}O]=\dfrac{n}{V}=\dfrac{x}{1L}=\dfrac{0.4mol}{1L}=0.4M
So, the equilibrium concentration of H2O{{H}_{2}}O is 0.4M.

Note: It should be noted that while writing the equilibrium expression of a reaction, the concentration of water molecules is not included if it is used as a solvent since it is a pure liquid.
However, in this question, since the water molecule is being produced as a result of the reaction between hydrogen gas and carbon dioxide, its concentration needs to be included in the equilibrium expression.