Question

The equilibrium constant ${K_P}$ for the reaction ${H_2}(g) + C{O_2}(g) \rightleftharpoons {H_2}O(g) + CO(g)$ is $4 \cdot 0$ at $1660^\circ C$. Initially $0 \cdot 80$ mole $C{O_2}$ are injected into a $5 \cdot 0$ litre flask. What is the equilibrium concentration of $C{O_2}(g)$:
A. $0 \cdot 533M$
B. $0 \cdot 0534M$
C. $5 \cdot 34M$
D. None of these

Answer 1

In the question above we need to find out the equilibrium concentration of carbon dioxide. As we know equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium.

Formula used:
${K_C} = \dfrac{{{{\left[ C \right]}^c} \times {{\left[ D \right]}^d} \times \cdot \cdot \cdot }}{{{{\left[ A \right]}^a} \times {{\left[ B \right]}^b} \times \cdot \cdot \cdot }}$,
when equation is the form,
$aA + bB + \cdot \cdot \cdot \rightleftharpoons cC + dD + \cdot \cdot \cdot$
Here, $K$ is equilibrium constant, $A,B, \cdot \cdot \cdot$ are products, $C,D, \cdot \cdot \cdot$ are reactants.

Complete step-by-step answer:
Now, as a common example, $\left[ A \right]$ is the equilibrium concentration of $A$ in moles.
Again, as a common example, $a$ is the number of moles of $A$
Now that we know the formula that is used, lets again write down our equation so that it gets easy while we solve,
${H_2}(g) + C{O_2}(g) \rightleftharpoons {H_2}O(g) + CO(g)$
They have given in the question that initially $0 \cdot 8$ moles of hydrogen and $0 \cdot 8$ moles of carbon is injected into $5 \cdot 0$ litre flask. Now initially as we know hydrogen and carbon dioxide both have $0 \cdot 8$ moles each, so it’s written down as an initial, but the reactant side both water and carbon monoxide are zero.
Now at equilibrium, it becomes $0 \cdot 8 - x$ under hydrogen and carbon dioxide and it becomes $x$ in the reactant side in water and carbon monoxide. Now let us analyse the changes that have occurred from initial to equilibrium. After that, we will proceed further.
Now as they have only asked us to find out the equilibrium concentration of carbon, and in the question they have given the value of equilibrium constant as $4$in$1660^\circ C$, so therefore the equation becomes,
${K_p} = 4 = \dfrac{{{x^2}}}{{{{\left( {0 \cdot 8 - x} \right)}^2}}}$
Now on solving the question, it becomes,
$2\left( {0 \cdot 8 - 2x} \right) = x$
This becomes,
$1 \cdot 6 = 3x$
So, therefore the value of $x$ becomes,
$x = \dfrac{{1 \cdot 6}}{3} = 0 \cdot 533$
Now before finding out equilibrium concentration, we need to find moles of carbon dioxide,
$= 0 \cdot 8 - 0 \cdot 533 \\\ = 0 \cdot 267M \\\$
Now it becomes easy to find out the equilibrium concentration of carbon dioxide,
$= \dfrac{{0 \cdot 267M}}{{5L}}$
Therefore the value becomes,
$= 0 \cdot 0534mol/L$

Hence, the right option is (B).

Note: Now equilibrium constants have different properties, every reaction has a particular unique value of equilibrium constant on a given temperature, they do not get affected by the initial concentration of the reactant, also with the change in temperature it changes.