Question

At 27\[^{0}C\] and 1 atmosphere pressure \[{{N}_{2}}{{O}_{4}}\] is 20% dissociated into \[N{{O}_{2}}\]. Find \[{{K}_{p}}\].
(A) 0.2
(B) 0.166
(C) 0.15
(D) 0.10

Answer 1

Hint: Note down the temperature and pressure of the surrounding provided in the question. Write down the complete dissociation reaction with the stoichiometric coefficients. Determine the equilibrium constant for the dissociation reaction. Now you can substitute the values obtained in the formula given below to find the equilibrium pressure.
\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\]
Where,
\[{{K}_{p}}\] is the equilibrium pressure constant
\[{{K}_{c}}\] is the equilibrium concentration constant
R is the universal gas constant
T is the temperature of the surrounding
\[\Delta n\] is the difference of number of moles of gaseous products and number of moles of gaseous reactants

Complete step by step solution:
Let us write the balanced reaction of the dissociation of ${{N}_{2}}{{O}_{4}}$ giving $N{{O}_{2}}$.
${{N}_{2}}{{O}_{4}}(g)\to \ \ 2N{{O}_{2}}(g)$
The formula for equilibrium constant is given below.
Formula: ${{\text{K}}_{\text{c}}}\text{=}\dfrac{\text{Concentration }{{\text{}}_{{}}}\text{ }{{\text{of }}_{{}}}\text{products }}{\text{Concentration }{{\text{ }}_{{}}}\text{o}{{\text{f }}_{{}}}\text{reactants }}$
For the above reaction, the equilibrium constant becomes,
${{\text{K}}_{\text{c}}}\text{=}\dfrac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}$
Since 20% of ${{N}_{2}}{{O}_{4}}$ only dissociates to give $N{{O}_{2}}$, we can say that the amount of $N{{O}_{2}}$ is 0.2 and the amount of ${{N}_{2}}{{O}_{4}}$ is 0.8 when the initial amount is taken as 1.
We will now substitute these values in the formula for equilibrium constant.
${{\text{K}}_{\text{c}}}\text{=}\dfrac{{{[0.2]}^{2}}}{[0.8]}$
\[\Delta n\] for the dissociation reaction is (2-1) = 1.
We will now substitute these values in the formula given in the hint to determine the value of \[{{K}_{p}}\].
\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\]
\[{{K}_{p}}=(0.06)\ \text{x }{{(0.0821\text{ x 300})}^{1}}\]
\[{{K}_{p}}\] = 0.166

Therefore, the value of \[{{K}_{p}}\] 0.166 atm. The correct answer is option (C).

Note: It is important to know that the concentration of reactant/product is raised to the power of its stoichiometric coefficient before substituting it into the equation for dissociation constant.