PCl5, PCl3 and Cl2 are at equilibrium at 500 K in a closed container and their concentrations are 0.8 × 10-3 mol L-1, 1.2 × 10-3 mol L-1 and 1.2 × 10-3 mol L-1 respectively. The value of Kc for the reaction: PCl5(g) leftharpoons PCl3(g) +Cl2(g) will be

Question

PCl5, PCl3 and Cl2 are at equilibrium at 500 K in a closed container and their concentrations are 0.8 × 10-3 mol L-1, 1.2 × 10-3 mol L-1 and 1.2 × 10-3 mol L-1 respectively. The value of Kc for the reaction: PCl5(g) leftharpoons PCl3(g) +Cl2(g) will be (A) 1.8×103 mol L-1 (B) 1.8×10-3 L mol-1 (C) 1.8×103 L mol-1 (D) 0.55 ×104 L mol-1

Tardigrade Admin · Accepted Answer

For the reaction: PCl5(g) rightleftharpoons PCl3(g) +Cl2(g) Kc=([PCl3][Cl2]/[PCl5]) =((1.2×10-3)2/0.8×10-3) =1.8×10-3

$1.8 \times 1 0^{3} m o l L^{- 1}$

$1.8 \times 1 0^{- 3} L m o l^{- 1}$

$1.8 \times 1 0^{3} L m o l^{- 1}$

$0.55 \times 1 0^{4} L m o l^{- 1}$

For the reaction :
$PC l_{5 (g)} r i g h tl e f t ha r p oo n s PC l_{3 (g)} + C l_{2 (g)}$
$K_{c} = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]}$
$= \frac{( 1.2 \times 1 0 ^{- 3} ) ^{2}}{0.8 \times 1 0 ^{- 3}}$
$= 1.8 \times 1 0^{- 3}$

Q. PCl5​, PCl3​ and Cl2​ are at equilibrium at 500K in a closed container and their concentrations are 0.8×10−3molL−1, 1.2×10−3molL−1 and 1.2×10−3molL−1 respectively. The value of Kc​ for the reaction : PCl5(g)​⇌PCl3(g)​+Cl2(g)​ will be

1.8×103molL−1

1.8×10−3Lmol−1

1.8×103Lmol−1

0.55×104Lmol−1

For the reaction : PCl5(g)​rightleftharpoonsPCl3(g)​+Cl2(g)​ Kc​=[PCl5​][PCl3​][Cl2​]​ =0.8×10−3(1.2×10−3)2​ =1.8×10−3

$1.8 \times 1 0^{3} m o l L^{- 1}$

$1.8 \times 1 0^{- 3} L m o l^{- 1}$

$1.8 \times 1 0^{3} L m o l^{- 1}$

$0.55 \times 1 0^{4} L m o l^{- 1}$

For the reaction :
$PC l_{5 (g)} r i g h tl e f t ha r p oo n s PC l_{3 (g)} + C l_{2 (g)}$
$K_{c} = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]}$
$= \frac{( 1.2 \times 1 0 ^{- 3} ) ^{2}}{0.8 \times 1 0 ^{- 3}}$
$= 1.8 \times 1 0^{- 3}$