Question

In equilibrium
${N_2}(g) + {O_2}(g) \rightleftharpoons 2NO(g),{K_C} = {10^{ - 40}}$
Initially 0.4 moles of ${N_2}$and 0.9 moles of ${O_2}$ were mixed, then moles of ${N_2}$,${O_2}$and $NO$at equilibrium are(approximately):
$1)0.4,0.9,6 \times {10^{ - 21}}$
$2)4 \times {10^{ - 21}},6 \times {10^{ - 21}},6 \times {10^{ - 21}}$
$3)0.4,0.9,0.9$
$4)6 \times {10^{ - 21}},0.4,0.9$

Answer 1

Hint: we know that equilibrium constant${K_c}$is the ratio of concentration of products and the concentration of reactions each raised to its stoichiometric coefficient. Thus, If know the value of equilibrium constant${K_c}$, we can employ it to form the linear equation in one variable since all the reactant species are interdependent and also the formed product is dependent on the quantity of the reactant, and thus solving that variable and getting the concentration of each molecule involved in the reaction.

Complete answer:
First of all, we need to understand the concentration conditions which prevail at the start of the reaction, i.e. time \[t = 0\]and then at equilibrium.
Considering the reaction:
                                           ${N_2}(g){\text{ }} + {\text{ }}{O_2}(g){\text{ }} \rightleftharpoons {\text{ }}2NO(g)$
At the start, \[t = 0\] 0.4mol 0.9mol 0 mol
At equilibrium,$t = {t_{eq}}$ $(0.4 - x){\text{ mol}}$ $(0.9 - x)mol$ $2x{\text{ mol}}$
The following formulation is because one mole of nitrogen gas and one mole of oxygen gas gives two moles of nitrogen monoxide.
Now, we know that ${K_c} = \dfrac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}}$
Substituting the values from above,we get \[{K_c} = \dfrac{{{{(2x)}^2}}}{{(0.4 - x){\text{ }}(0.9 - x)}}\]
Since, in the question, we are given the value of ${K_c}$=${10^{ - 40}}$. Hence substituting it into the formed equation, we get:
\[{10^{ - 40}} = \dfrac{{{{(2x)}^2}}}{{(0.4 - x){\text{ }}(0.9 - x)}}\]
Now we know that x will be very small and hence $0.4 - x \approx 0.4$ and $0.9 - x \approx 0.9$. Thus, substituting them again, we get
\[{10^{ - 40}} = \dfrac{{{{(2x)}^2}}}{{(0.4){\text{ }}(0.9)}}\]
=\[{10^{ - 40}} \times (0.4){\text{ }} \times (0.9) = {(2x)^2}\]
Thus, we get: $4{x^2} = 0.36 \times {10^{ - 40}}$
Solving this equation, we get the value of $x = 0.3 \times {10^{ - 20}}$, which is equal to $x = 3 \times {10^{ - 21}}$
Now, with the value of $x$ known, we can find the concentration of all the species involved in the reaction at the equilibrium
Concentration of ${N_2}$ at equilibrium=$(0.4 - x){\text{ mol}}$ =$0.4 - 3 \times {10^{ - 21}} \approx 0.4$mol
Concentration of ${O_2}$ at equilibrium=$(0.9 - x)mol$=$0.9 - 3 \times {10^{ - 21}} \approx 0.9$mol
Concentration of at $NO$equilibrium=$2x{\text{ mol}}$=$2 \times 3 \times {10^{ - 21}} = 6 \times {10^{ - 21}}$mol

So, the correct answer is “Option 1”.

Note:
 It is important to understand the role of stoichiometry in the reaction equilibrium conditions. The number of moles of products formed by how many moles of reactants, that plays a crucial role in the determination of the concentration as well as calculation. Also, since these reactants are interdependent and product amount depends on the reactants’ amount, it is important to balance the reaction first and then solve.