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For a reaction carried out at 25°C with an equilibrium constant of 1 × 10−3, in order to increase the equilibrium constant by a factor of 10:

a. How much must Δ³Ò° change?

b. How much must Δ±á° change if Δ³§Â° = 0 kcal mol-1 K-1?

c. How much must Δ³§Â° change if Δ±á° = 0 kcal mol-1?

Short Answer

Expert verified

a) Δ³Òo = _RT ln Keq

= - 1.36 kcal/mol

²ú)Δ³Òo = Δ±áo - T Δ³§o

Δ±áo= -1.36 kcal/mol

³¦)Δ³Òo = Δ±áo - T Δ³§o

Δ³§o= 4.56 × 10-3 kcal/(mol deg)

Step by step solution

01

Some definition

What is Δ³§Â° value =

S value is always positive. But Δ³§Â° denotes change in entropy. Entropy means the degree of randomness and disorder of the molecules. It can be negative or positive. If the products are less than the reactants then disorder decreases so Δ³§Â° value becomes negative. If the reactant and products are in same number then the Δ³§Â° value becomes zero. If the products are more in number than the reactant then Δ³§Â° value will be positive.

What is Δ±á° value =

A positive Δ±á°value value represents addition of energy from the reaction then the reaction is endothermic reaction. A negative value represents removal of energy from the reaction then the reaction is exothermic reaction.

What is equilibrium constant=

It provides relationship between the products and reactants when a chemical reaction reaches equilibrium.

02

Final answer

Given:

T = 25°C = 298 K

Keq = 1 × 10−3

a)We know,

Δ³Òo = _RT ln Keq

= - 1.986 × 10-3 × 298 K × ln (10)

= - 1.986 × 10-3 × 298 K × 2.3

= - 1.36 kcal/mol

b) Now, we will have to calculate Δ Ho, from the given data by using the formula

Δ³Òo = Δ±áo - T Δ³§o

-1.36 = Δ±áo – 0

Δ±áo= -1.36 kcal/mol

c) Now, we will have to calculate Δ So, from the given data by using the formula

Δ³Òo = Δ±áo - T Δ³§o

-1.36 = 0 – 298 Δ³§o

Δ³§o= 4.56 × 10-3 kcal/(mol deg)

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