Question

At 540 K, $0.10$ mole of $PC{l}_5$ are heated in a $8$ litre flask . The pressure of the equilibrium mixture is found to be $1.0$ atm .

Calculate $K_p$ and $K_c$ for the reaction.

• A. $4\times {10}^{-2}$ mole lit${}^{-1}$ and $1.77$ atm
• B. $0.4\times {10}^{-2}$ mole lit${}^{-1}$ and $0.177$ atm
• C. $8\times {10}^{-2}$ mole lit${}^{-1}$ and $0.177$ atm
• D. $4\times {10}^2$ mole lit${}^{-1}$ and $1.77$atm

Answer 1

The correct option is A $4\times {10}^{-2}$ mole lit${}^{-1}$ and $1.77$ atm

The equilibrium reaction is $PC{l}_5\left(g\right)\iff PC{l}_3\left(g\right)+C{l}_2\left(g\right)$.
Following table lists quantites of various species.

	$PC{l}_5$	$PC{l}_3$	$C{l}_2$
Initial	0.10	0	0
Equilibrium	$0.1-x$	$x$	$x$

But $PV=nRT$ and total pressure is 1 atm and volume is 8 L.

Hence, $1\times 8=(0.1+x)\times 0.082\times 540$.
Hence, $x=0.08$.
Hence, the equilibrium number of moles of $PC{l}_5,PC{l}_3$ and $C{l}_2$ are (0.1-0.08), 0.08 and 0.08 respectively.
The expression for the equilibrium constant is $K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}=\frac{0.08\times 0.081}{8(0.1-0.08)}=4\times {10}^{-2}mollitr{e}^{-1}$.
$\Delta n=1$. Hence, $K_p=K_c\left(RT\right)=4\times {10}^{-2}(0.082\times 540)=1.77atm$.