PCl5, PCl3 and Cl2 are at equilibrium at 500 K with concentration 2.1 M PCl3, 2.1 M Cl2 and 1.9 M PCl5. The equilibrium constant for the given reaction is PCl5(g) leftharpoons PCl3(g) + Cl2(g)

Question

PCl5, PCl3 and Cl2 are at equilibrium at 500 K with concentration 2.1 M PCl3, 2.1 M Cl2 and 1.9 M PCl5. The equilibrium constant for the given reaction is PCl5(g) leftharpoons PCl3(g) + Cl2(g) (A) 2.32 (B) 1.79 (C) 4.2 (D) 3.8

Tardigrade Admin · Accepted Answer

Kc=([PCl3][Cl2]/[PCl5]) =(2.1×2.1/1.9) =2.32

Q. $PC l_{5}$ , $PC l_{3}$ and $C l_{2}$ are at equilibrium at $500 K$ with concentration $2.1 M PC l_{3}$ , $2.1 M C l_{2}$ and $1.9 M PC l_{5}$ . The equilibrium constant for the given reaction is
$PC l_{5 (g)} ⇌ PC l_{3 (g)} + C l_{2 (g)}$

$2.32$

$1.79$

$4.2$

$3.8$

$K_{c} = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]}$
$= \frac{2.1 \times 2.1}{1.9}$
$= 2.32$

Q. PCl5​, PCl3​ and Cl2​ are at equilibrium at 500K with concentration 2.1MPCl3​, 2.1MCl2​ and 1.9MPCl5​. The equilibrium constant for the given reaction is PCl5(g)​⇌PCl3(g)​+Cl2(g)​

2.32

1.79

4.2

3.8