Question

At $30^\circ {\text{C}}$ , the gaseous mixture for the equilibrium, ${{\text{N}}_{\text{2}}}\left( {\text{g}} \right) + {{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \Leftrightarrow 2{\text{NO}}\left( {\text{g}} \right)$ , contains 56 gm of ${{\text{N}}_{\text{2}}}$ , 128 gm of ${{\text{O}}_{\text{2}}}$ and 120 gm of ${\text{NO}}$ at one atmospheric pressure. The value of ${{\text{K}}_{\text{p}}}$ at $30^\circ {\text{C}}$ is:
A.$\dfrac{1}{2}$
B.$2$
C.$1$
D.$\dfrac{1}{4}$

Answer 1

According to the law of chemical equilibrium, the product of the molar concentrations of the product species, each raised to the power equal to its coefficient divided by the product of the molar concentrations of the reactant species, each raised to the power equal to its coefficient at constant temperature is constant. This constant is called the equilibrium constant.
If we consider a general reaction at the state of equilibrium
${\text{aA + bB}}\overset {} \leftrightarrows {\text{cC + dD}}$

Complete step by step answer:
The value of the equilibrium constant, ${{\text{K}}_{\text{c}}}$ can be written as:
${{\text{K}}_{\text{c}}} = \left( {\dfrac{{{{\left[ {\text{C}} \right]}^{\text{c}}}{{\left[ {\text{D}} \right]}^{\text{d}}}}}{{{{\left[ {\text{A}} \right]}^{\text{a}}}{{\left[ {\text{B}} \right]}^{\text{b}}}}}} \right)$
This expression represents the law of chemical equilibrium.
For gas phase reactions, the equilibrium constant can also be expressed in terms of partial pressures of the reactants and products and in that case, the equilibrium constant is denoted by ${{\text{K}}_{\text{p}}}$ .
Complete step by step answer:
Given that at $30^\circ {\text{C}}$ , the gaseous mixture for the equilibrium, ${{\text{N}}_{\text{2}}}\left( {\text{g}} \right) + {{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \Leftrightarrow 2{\text{NO}}\left( {\text{g}} \right)$ , contains 56 gm of ${{\text{N}}_{\text{2}}}$ , 128 gm of ${{\text{O}}_{\text{2}}}$ and 120 gm of ${\text{NO}}$ at one atmospheric pressure.
We need to find out the value of ${{\text{K}}_{\text{p}}}$ at $30^\circ {\text{C}}$ .
For the general reaction in gas phase${\text{aA + bB}}\overset {} \leftrightarrows {\text{cC + dD}}$ we have :
${{\text{K}}_{\text{p}}} = \dfrac{{{{\text{p}}_{\text{C}}}^{\text{c}}{\text{.}}{{\text{p}}_{\text{D}}}^{\text{d}}}}{{{{\text{p}}_{\text{A}}}^{\text{a}}{\text{.}}{{\text{p}}_{\text{B}}}^{\text{b}}}}$
Here, ${{\text{p}}_{\text{A}}}{\text{,}}{{\text{p}}_{\text{B}}}{\text{,}}{{\text{p}}_{\text{C}}}$ and ${{\text{p}}_{\text{D}}}$ represent the equilibrium partial pressures of A, B, C and D respectively.
Let us first calculate the number of moles of the reactants and the products.
The mass of ${{\text{N}}_{\text{2}}}$ is given to be 56 gm and its molar mass is 28 gram per mole.
Therefore, the number of moles of ${{\text{N}}_{\text{2}}}$ will be $\dfrac{{56}}{{28}} = 2$ .
The mass of ${{\text{O}}_{\text{2}}}$ is 128 gm and its molar mass is 32 gram per mole. So, the number of moles of ${{\text{O}}_{\text{2}}}$ is $\dfrac{{128}}{{32}} = 4$ .
The mass of ${\text{NO}}$ is 120 gm and its molar mass is 30 gram per mole. So, the number of moles of ${{\text{O}}_{\text{2}}}$ is $\dfrac{{120}}{{30}} = 4$ .
So, the partial pressure of ${{\text{N}}_{\text{2}}}$ is $\dfrac{2}{{\left( {2 + 4 + 4} \right)}} \times 1 = 0.2$
The partial pressure of ${{\text{O}}_{\text{2}}}$ is $\dfrac{4}{{\left( {2 + 4 + 4} \right)}} \times 1 = 0.4$
The partial pressure of ${\text{NO}}$ is $\dfrac{4}{{\left( {2 + 4 + 4} \right)}} \times 1 = 0.4$
Therefore, we have

${{\text{K}}_{\text{p}}} = \dfrac{{{{\text{p}}_{{\text{NO}}}}^{\text{2}}}}{{{{\text{p}}_{{{\text{N}}_{\text{2}}}}}^{\text{1}}{\text{.}}{{\text{p}}_{{{\text{O}}_{\text{2}}}}}^{\text{1}}}} \\\ \Rightarrow {{\text{K}}_{\text{p}}} = \dfrac{{{{\left( {0.4} \right)}^2}}}{{0.2 \times 0.4}} \\\ \Rightarrow {{\text{K}}_{\text{p}}} = 2 \\\$

So, B is the correct option.

Note:
For some reactions the values of ${{\text{K}}_{\text{p}}}$ and ${{\text{K}}_{\text{c}}}$ are equal. But for many other reactions, they have different values and hence it is preferable to calculate one from the other.
The relation between ${{\text{K}}_{\text{c}}}$and ${{\text{K}}_{\text{p}}}$ is given by the following expression:
${{\text{K}}_{\text{c}}}{\text{ = }}{{\text{K}}_{\text{p}}} \times \dfrac{{\text{1}}}{{{{\left( {{\text{RT}}} \right)}^{\Delta {\text{n}}}}}}$
Here, $\Delta {\text{n}}$ represents the difference between the number of moles of gaseous products and that of gaseous reactants. The numerical value of $\Delta {\text{n}}$ is obtained from the coefficients of the balanced equation.
R and T are the gas constant and temperature respectively.