Question

The equilibrium constant $K_p$ (in atm) for the reaction is 9 at 7 atm and 300 K. $A_2\left(g\right)\rightleftharpoons B_2\left(g\right)+C_2\left(g\right)$. Calculate the average molar mass (in gm/mol)of an equilibrium mixture. Given: Molar mass of $A_2,B_2$ and $C_2$ are 70,49 &21 gm/mol respectively.

• A. $50$
• B. $45$
• C. $40$
• D. $37.5$

Answer 1

Answer: (A)

$50$

The relation between $K_p$ and $K_c$ is $K_p=K_c(RT{)}^{\Delta n}=K_c(RT)$ because $\Delta n=1$

$9=K_c\times 0.0821\times 300$

$K_c=0.3654$

Let the equilibrium number of moles (and molar concentration) of A,B and C be 1,x and x respectively.

$K_c=\frac{\left[A\right]\left[B\right]}{\left[C\right]}=\frac{X^2}{1}=0.3654$

$x=0.6044$

Thus equilibrium moles of A,B and C are 1,0.6 and 0.6 respectively.

The mole fractions of A,B and C are 0.455, 0.27 and 0.27 respectively.

Average molecular weight $=70\times 0.455+49\times 0.27+21\times 0.27=50$