Sulfuryl chloride is in equilibrium with sulfur dioxide and chlorine gas: so2cl2(g) so2(g) + cl2(g) a system with a volume of 1.00 l is in equilibrium at a certain temperature with p(so2cl2) = 1.00 bar and p(so2) = p(cl2) = 0.10 bar. by how much will the number of moles of so2cl2 at equilibrium change if the volume is reduced to 0.50 l? (a) increase 1-10% (b) increase 11-50% (c) decrease 1-10% (d) decrease 11-50%

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

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explanation:

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