Question

The equilibrium constant for the reaction $C{O_{(g)}} + {H_2}{O_{(g)}} \rightleftharpoons C{O_{2(g)}} + {H_{2(g)}}$ is ${\text{3}}$ at ${\text{500K}}$. In a ${\text{2 litre}}$ vessel ${\text{60 gm}}$ of water gas [equimolar mixture of $C{O_{(g)}}$ and ${H_{2(g)}}$] and ${\text{90 gm}}$ of steam is initially taken.
What is the concentration of ${H_{2(g)}}$ at equilibrium ${\text{(mole/L)}}$?
A) ${\text{1}} \cdot {\text{75}}$
B) ${\text{3}} \cdot {\text{5}}$
C) ${\text{1}} \cdot {\text{5}}$
D) ${\text{0}} \cdot {\text{75}}$

Answer 1

Hint:The weight of water is given and from that one can calculate the number of moles of mixture and water. One can take the value of carbon monoxide as ${\text{x}}$ and can calculate the concentration for all the components. The formula for equilibrium constant is the concentration of all products divided by the concentration of all reactants. One must remember that concentration of ${H_{2(g)}}$ is needed to calculate.

Complete answer:
1) First of all, we will calculate the number of moles of mixture and water as below,
The molecular weight of the gas mixture $ = CO + {H_2} = (12 + 16) + 2 = 30grams$
Number of moles of the gas mixture $ = \dfrac{{weight}}{{{\text{Molecular weight}}}} = \dfrac{{60}}{{30}} = 2{\text{ moles}}$
The molecular weight of water $ = {H_2}O = 2 + 16 = 18{\text{ grams}}$
Number of moles of water $ = \dfrac{{weight}}{{{\text{Molecular weight}}}} = \dfrac{{90}}{{18}} = 5{\text{ moles}}$
2) Now let us see the equation for the equilibrium as below,
$C{O_{(g)}} + {H_2}{O_{(g)}} \rightleftharpoons C{O_{2(g)}} + {H_{2(g)}}$
In the above reaction, as we have calculated the initial moles of ${\text{CO}}$ are ${\text{2 moles}}$ and for ${H_2}O$ they are ${\text{5 moles}}$ and for $C{O_2}$ let us consider as ‘x’ and for ${H_2}$ it is ${\text{2 moles}}$.
3) So, in the above reaction if we suppose ‘x’ moles $C{O_2}$ are formed then it will be formed from the initial moles of all reactants and we can take it as for ${\text{CO}}$ as ${\text{(2 - x)}}$, for ${H_2}$ as ${\text{(5 - x)}}$ , and for ${H_2}$ as ${\text{(2 + x)}}$. These values are of moles of individual reactants and products. Now let’s see the equation for the equilibrium as below,
${K_{eq{u^m}}} = \dfrac{{{\text{Concentration of products}}}}{{{\text{Concentration of reactants}}}}$
${K_{eq{u^m}}} = \dfrac{{\left[ {C{O_2}} \right]\left[ {{H_2}} \right]}}{{\left[ {CO} \right]\left[ {{H_2}O} \right]}}$
4) As we have calculated the number of moles, the concentration will be the number of moles divided by volume, and let’s put that in the above equation,
${K_{equ}} = \dfrac{{\dfrac{x}{2} \times \dfrac{{(2 + x)}}{2}}}{{\dfrac{{(2 - x)}}{2} + \dfrac{{(5 - x)}}{2}}}$
In the above equation, $\dfrac{1}{2}$ will gets cancelled and the value of ${K_{equ}}$ is given as ${\text{3}}$ then we get,
$3 = \dfrac{{x\left( {2 + x} \right)}}{{\left( {2 - x} \right) + \left( {5 - x} \right)}}$
Now by calculating the above equation we get the value of ${\text{x}}$ as,
$x = 1 \cdot 5$
5) As we need to calculate the concentration ${H_2}$, we first need moles of ${H_2}$ which are as below,
${H_2} = (2 + x) = (2 + 1 \cdot 5) = 3 \cdot 5{\text{ moles}}$
As we now know the number of moles of ${H_2}$ we can calculate the concentration as below,
$Concentration({H_2}) = \dfrac{{{\text{Number of moles}}}}{{Volume}}$
$\left[ {{H_2}} \right] = \dfrac{{3 \cdot 5}}{2} = 1 \cdot 75{\text{ mole/L}}$
Therefore, the equilibrium concentration ${H_2}$ at equilibrium is $1 \cdot 75{\text{ mole/L}}$ which shows option A as the correct choice.

Note:
The equilibrium is a state in a chemical reaction where the concentration of products and reactants will not be changed further with time. This concept can be used to determine the concentration of any reactant and product in the reaction.