Which reaction would cause an increase in entropy? a. so2cl2(g) so2(g) + cl2(g) b. pcl3(g) + cl2(g) pcl5(g) c. 2co(g) + o2(g) 2co2(g) d. c2h2(g) + h2(g) c2h4(g)

A)

Explanation: Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

A)

1 mole of gaseous reactant is converting to 2 gaseous products, thus the entropy increases.

B)

2 moles of gaseous reactants are converting to 1 gaseous products, thus the entropy decreases.

C)

3 moles of gaseous reactants are converting to 1 gaseous product, thus the entropy decreases.

D.

2 moles of gaseous reactants are converting to 1 gaseous product, thus the entropy decreases.

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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