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Question: \(\mathrm { PCl } _ { 5 } , \mathrm { PCl } _ { 3 }\) and \(\mathrm { Cl } _ { 2 }\) are at equilib...

PCl5,PCl3\mathrm { PCl } _ { 5 } , \mathrm { PCl } _ { 3 } and Cl2\mathrm { Cl } _ { 2 } 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−10.8 \times 10 ^ { - 3 } \mathrm {~mol} \mathrm {~L} ^ { - 1 } , 1.2 \times 10 ^ { - 3 } \mathrm {~mol} \mathrm {~L} ^ { - 1 } and

1.2×10−3 mol L−11.2 \times 10 ^ { - 3 } \mathrm {~mol} \mathrm {~L} ^ { - 1 } respectively.

PCl3( g)+Cl3( g)\mathrm { PCl } _ { 3 ( \mathrm {~g} ) } + \mathrm { Cl } _ { 3 ( \mathrm {~g} ) } will be.

A

1.8×103molL−11.8 \times 10 ^ { 3 } \mathrm { molL } ^ { - 1 }

B

1.8×10−31.8 \times 10 ^ { - 3 }

C

1.8×10−3 L mol−11.8 \times 10 ^ { - 3 } \mathrm {~L} \mathrm {~mol} ^ { - 1 }

D

0.55×1040.55 \times 10 ^ { 4 }

Answer

1.8×10−31.8 \times 10 ^ { - 3 }

Explanation

Solution

: For the reaction:,

Kc=[PCl3][Cl2][PCl5]=(1.2×10−3)20.8×10−3=1.8×10−3\mathrm { K } _ { \mathrm { c } } = \frac { \left[ \mathrm { PCl } _ { 3 } \right] \left[ \mathrm { Cl } _ { 2 } \right] } { \left[ \mathrm { PCl } _ { 5 } \right] } = \frac { \left( 1.2 \times 10 ^ { - 3 } \right) ^ { 2 } } { 0.8 \times 10 ^ { - 3 } } = 1.8 \times 10 ^ { - 3 }