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When the reaction A+B⇔C is at equilibrium, the concentrations of reactants are as follows: A=2mM,B=3mM,C=9mM. What is the standard free energy change for the reaction?

Short Answer

Expert verified

The standard free energy change is Δ³Ò0=-18122.29J/mol.

Step by step solution

01

Given data

The given reaction is,

A+B⇔C

Δ³Ò0=-18122.29J/mol

A and Bare the substrates and C is the product.

Then,

A+B⇔CA=0.002MB=0.003MC=0.009MA=0.002MB=0.003MC=0.009M

02

Calculating the equilibrium constant 

Keq'=CA×B=0.009M0.002M×0.003M=1500M

03

Calculating the standard free energy change 

ΔG=ΔG0+RTlnQ

Where,

Δ³ÒChange in free energy

ΔG0Standard free energy change

R gas constant

T Standard temperature

04

At equilibrium

ΔG0=-RTlnKeq'=-2.303×8.314JK-1mol-1×298K×ln1500M=-18122.29J/mol

Therefore, the standard free energy change isΔ³Ò0=-18122.29J/mol.

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