Question

For reaction, $PCl_{3}\left(\right.g\left.\right)+Cl_{2}\left(\right.g\left.\right)\rightleftharpoons PCl_{5}\left(\right.g\left.\right)$ , the value of $K_{c}$ at $250^\circ C$ is 26. The value of $K_{p}$ at this temperature will be

• A. 0.61
• B. 0.57
• C. 0.83
• D. 0.46

Answer 1

Answer: (A)

$K _{ p }= K _{ c }( RT )^{\Delta n }$

Here $K _{ c }=26, R =0.0821 L$ atm $mol ^{-1} K ^{-1},$

$T =250+273=523 K$

$\Delta n =1-(1+1)=-1$

So $K _{ p }=26 \times(0.0821 \times 523)^{-1}=0.605 \simeq 0.61$