Question

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

The correct option is A $50$

$Kp=Kc(RT{)}^{\Delta n}$

since $\Delta n=1$, then $Kc\left(RT\right)$

$9=Kc\times 0.0821\times 300$

$Kc=0.3654$

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

$x=0.6044$

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

$\therefore \left(A\right)$ is correct option