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Question: For a reaction, \(2 \mathrm { SO } _ { 2 ( \mathrm {~g} ) } + \mathrm { O } _ { 2 ( \mathrm {~g} ) }...

For a reaction, 2SO2( g)+O2( g)2 \mathrm { SO } _ { 2 ( \mathrm {~g} ) } + \mathrm { O } _ { 2 ( \mathrm {~g} ) } 2SO3( g),1.52 \mathrm { SO } _ { 3 ( \mathrm {~g} ) } , 1.5 moles of SO2\mathrm { SO } _ { 2 } and 1 mole of O2\mathrm { O } _ { 2 } are taken in a 2 L vessel. At equilibrium the concentration of SO3\mathrm { SO } _ { 3 } was found to be 0.35molL−10.35 \mathrm { molL } ^ { - 1 }The for the reaction would be

A
B
C

0.6 L mol−10.6 \mathrm {~L} \mathrm {~mol} ^ { - 1 }

D

2.95 L mol−12.95 \mathrm {~L} \mathrm {~mol} ^ { - 1 }

Answer
Explanation

Solution

: 2SO2( g)+2 \mathrm { SO } _ { 2 ( \mathrm {~g} ) } + 2SO3( g)2 \mathrm { SO } _ { 3 ( \mathrm {~g} ) }

Initial conc. 1.52\frac { 1.5 } { 2 } 12\frac { 1 } { 2 } 0

At equilibrium 1.52−0.35\frac { 1.5 } { 2 } - 0.35 12−0.35\frac { 1 } { 2 } - 0.35 0.350.35

=0.4= 0.4 =0.15= 0.15 0.350.35

Kc=(0.35)20.4×0.4×0.15=5.1 L mol−1\mathrm { K } _ { \mathrm { c } } = \frac { ( 0.35 ) ^ { 2 } } { 0.4 \times 0.4 \times 0.15 } = 5.1 \mathrm {~L} \mathrm {~mol} ^ { - 1 }