Solveeit Logo
Login

Question

Question: In equilibrium \({N_2}(g) + {O_2}(g) \rightleftharpoons 2NO(g),{K_C} = {10^{ - 40}}\) Initially ...

In equilibrium
N2(g)+O2(g)⇌2NO(g),KC=10−40{N_2}(g) + {O_2}(g) \rightleftharpoons 2NO(g),{K_C} = {10^{ - 40}}
Initially 0.4 moles of N2{N_2}and 0.9 moles of O2{O_2} were mixed, then moles of N2{N_2},O2{O_2}and NONOat equilibrium are(approximately):
1)0.4,0.9,6×10−211)0.4,0.9,6 \times {10^{ - 21}}
2)4×10−21,6×10−21,6×10−212)4 \times {10^{ - 21}},6 \times {10^{ - 21}},6 \times {10^{ - 21}}
3)0.4,0.9,0.93)0.4,0.9,0.9
4)6×10−21,0.4,0.94)6 \times {10^{ - 21}},0.4,0.9

Explanation

Solution

we know that equilibrium constantKc{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 constantKc{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=0t = 0and then at equilibrium.
Considering the reaction:
N2(g) + O2(g) ⇌ 2NO(g){N_2}(g){\text{ }} + {\text{ }}{O_2}(g){\text{ }} \rightleftharpoons {\text{ }}2NO(g)
At the start, t=0t = 0 0.4mol 0.9mol 0 mol
At equilibrium,t=teqt = {t_{eq}} (0.4−x) mol(0.4 - x){\text{ mol}} (0.9−x)mol(0.9 - x)mol 2x mol2x{\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 Kc=[NO]2[N2][O2]{K_c} = \dfrac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}}
Substituting the values from above,we get Kc=(2x)2(0.4−x) (0.9−x){K_c} = \dfrac{{{{(2x)}^2}}}{{(0.4 - x){\text{ }}(0.9 - x)}}
Since, in the question, we are given the value of Kc{K_c}=10−40{10^{ - 40}}. Hence substituting it into the formed equation, we get:
10−40=(2x)2(0.4−x) (0.9−x){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≈0.40.4 - x \approx 0.4 and 0.9−x≈0.90.9 - x \approx 0.9. Thus, substituting them again, we get
10−40=(2x)2(0.4) (0.9){10^{ - 40}} = \dfrac{{{{(2x)}^2}}}{{(0.4){\text{ }}(0.9)}}
=10−40×(0.4) ×(0.9)=(2x)2{10^{ - 40}} \times (0.4){\text{ }} \times (0.9) = {(2x)^2}
Thus, we get: 4x2=0.36×10−404{x^2} = 0.36 \times {10^{ - 40}}
Solving this equation, we get the value of x=0.3×10−20x = 0.3 \times {10^{ - 20}}, which is equal to x=3×10−21x = 3 \times {10^{ - 21}}
Now, with the value of xx known, we can find the concentration of all the species involved in the reaction at the equilibrium
Concentration of N2{N_2} at equilibrium=(0.4−x) mol(0.4 - x){\text{ mol}} =0.4−3×10−21≈0.40.4 - 3 \times {10^{ - 21}} \approx 0.4mol
Concentration of O2{O_2} at equilibrium=(0.9−x)mol(0.9 - x)mol=0.9−3×10−21≈0.90.9 - 3 \times {10^{ - 21}} \approx 0.9mol
Concentration of at NONOequilibrium=2x mol2x{\text{ mol}}=2×3×10−21=6×10−212 \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.