Question

The equilibrium constant $K_p$( in atm) for the reaction is $2.46$ at $7$ atm and $300\text{K}$.
$A_2\left(g\right)\rightleftharpoons B_2\left(g\right)+C_2\left(g\right)$
Calculate the average molar mass (in g/mol) of an equilibrium mixture.
Given: Molar mass of $A_2,B_2$and $C_2$ are $70,49$ and $21\text{g/mol}$ respectively.
(Given $R=0.082{\text{L atm K}}^{-1}{\text{mol}}^{-1},\sqrt{0.1}\approx 0.3$

• A. $56.7$
• B. $45$
• C. $40$
• D. $37.5$

Answer 1

The correct option is A $56.7$

A) Given $K_p=2.46$
$R=0.082{\text{L atm K}}^{-1}{\text{mol}}^{-1}$
$P=7\text{atm}$
$T=300\text{K}$
$\Delta n_g=$Number of moles of Product-Number of moles of reactant $=2-1=1$
$\therefore K_p=K_c(RT{)}^{△n_g}$
$2.46=K_c(0.082\times 300{)}^1$
$\Rightarrow K_c=\frac{2.46}{0.082\times 300}=0.1$
Let equilibrium number of moles of A,B and C be $1,x$ and $x$ respectively.
$K_c=\frac{\left[B_2\right]\left[C_2\right]}{\left[A_2\right]}$
On substituting all values, we get
$0.1=\frac{\left(x\right)\times \left(x\right)}{1}$
$\Rightarrow X^2=0.1$
$\Rightarrow X=0.3$
So. equilibrium moles of A, B and C are $1,0.3$ and $0.3$ respectively.

Mole fraction of $A\left({\chi }_A\right)=\frac{n_{A_2}}{n_{A_2}+n_{B_2}+n_{C_2}}$

$=\frac{1}{1+0.3+0.3}=\frac{1}{1.6}=0.625\approx 0.62$

Similarly ${\chi }_{B_2}=\frac{0.3}{1.6}=0.1875\approx 0.19$

$\therefore n_{B_2}=n_{C_2}$

$\because {\chi }_{B_2}={\chi }_{C_2}=0.19$

Average molecular mass,
$=M_{A_2}\times {\chi }_{A_2}+M_{B_2}\times {\chi }_{B_2}+M_{C_2}\times {\chi }_{C_2}$
$=70\times 0.62+49\times 0.19+21\times 0.19$
$=56.7$

$M_{avg}=56.7\text{g/mol}$
Correct option is (A).