Question

If 1.0 mol each of all the four gases is taken in a one litre container the concentration of $N{O_2}$ at equilibrium would be:
A.$1.6mol{L^{ - 1}}$
B.$0.8mol{L^{ - 1}}$
C.$0.4mol{L^{ - 1}}$
D.$0.6mol{L^{ - 1}}$

Answer 1

We should know if the reaction needs to be balanced in such cases. Also, the terms ${k_p},{k_c}$ should be clear to us. Few key points in this question that we need to learn are:
${k_c}$ , the equilibrium constant is independent of the initial concentration of the reactants.
${k_c}$varies with variation in temperature.
${k_c}$depends upon the nature of reaction
Formula used: In terms of molarity, Equilibrium constant, ${k_c} = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}$

Complete step by step answer:
Before approaching the solution, we must understand the basics required to tackle the question.
Consider the following reaction:
$aA + bB \rightleftharpoons cC + dD$-(1)
A and B are the reactants. C and D are the products formed. $a,b,c,d$ represents the number of moles of the components. is the symbol that shows that equilibrium exists.
In a reversible chemical reaction, Equilibrium constant is defined as the value that depicts the relationship between the amount (in the term of molarity, pressure, or concentration) of components (products and reactants) existing in equilibrium at a given temperature.
For equation (1)
$S{O_2}\left( g \right) + N{O_2}\left( g \right) \rightleftharpoons S{O_3}\left( g \right) + NO\left( g \right)$
Equilibrium constant in terms of partial pressure, ${k_p} = \dfrac{{{{\left( C \right)}^c}{{\left( D \right)}^d}}}{{{{\left( A \right)}^a}{{\left( B \right)}^b}}}$
Equilibrium constant in terms of molarity, ${k_c} = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}$
We need to take the second equation into consideration.
For the given equation,
$S{O_2}\left( g \right) + N{O_2}\left( g \right) \rightleftharpoons S{O_3}\left( g \right) + NO\left( g \right)$ The equation is balanced already.
${k_c} = 16$
Initial concentration of all four gases is 1$M$each (given in the question)

Name of the gas	Equilibrium concentration(in terms of molarity$M$)
$S{O_2}$(reactant)	$1 - x$
$N{O_2}$(reactant)	$1 - x$
$S{O_2}$(product)	$1 + x$
$NO$(product)	$1 + x$

Using equilibrium constant in terms of molarity, We get
${k_c} = \dfrac{{\left[ {S{O_2}} \right]\left[ {NO} \right]}}{{\left[ {S{O_2}} \right]\left[ {\left[ {N{O_2}} \right]} \right]}}$
Substituting the values from the above table
$16 = \dfrac{{\left( {1 + x} \right)\left( {1 + x} \right)}}{{\left( {1 - x} \right)\left( {1 - x} \right)}}$
Solving for $x$,

$$16 = \dfrac{{{{\left( {1 + x} \right)}^2}}}{{{{\left( {1 - x} \right)}^2}}} \\\ 4 = \dfrac{{1 + x}}{{1 - x}} \\\ 4\left( {1 - x} \right) = 1 + x \\\ 4 - 4x = 1 + x \\\ 3 = 5x \\\ x = 0.6 \\\$$

So, at the equilibrium, the concentration of $\left[ {N{O_2}} \right] = 1 - x = 1 - 0.6 = 0.4mol{L^{ - 1}}$

Thus, option C. is the correct option for the given question.

Note:
One should avoid any kind of calculation mistakes so that solution to the question does not need any kind of review. We should revise the basic concept for the long run because it will help us in making good hold on such concepts from the exam point of view.