Question

imagine that for the unimolecular reaction A ⟺ B, the value of Δ G° is -0.89 kcal/mol. If equimolar initial amounts of A and B react at 25 °C, what fraction of the molecules will be B molecules once the system has come to equilibrium? Report your answer as a percent to the nearest ones.

Answer 1

Consider a reaction A B .

The equilibrium constant K = [ B ] / [ A ]

We know that , - G ⁰ = 2.303 R T log K

   - ( - 0.89 kcal / mol ) = 2.303 x 1.985 x 10 ^-3 kcal K ^{- 1} mol ^{- 1} x 298.15 K x log K

0.89 kcal / mol = 2.303 x 1.985 x 10 ^-3 kcal K ^{- 1} mol ^{- 1} x 298.15 K x log K

   log K= 0.89 kcal / mol / 2.303 x 1.985 x 10 ^-3 kcal K ^{- 1} mol ^{- 1} x 298.15 K

= 0.653

K = 4.50

Calculate Q c of reaction to predict direction of reaction .

The equilibrium constant K = [ B ] / [ A ]  

Given [ B ] = [ A ]

Therefore, Q c = 1

Q c < K. Reaction will shift to the right .

Let's use ICE table.

Concentration	A	B
Initial	1	1
Change	- X	+X
Equilibrium	1- X	1 + X

K = [ B ] / [ A ] = 1 + X / 1 - X = 4.50

1 + X = 4.50 ( 1- X )

= 4.5 - 4.5 X

X + 4.5 X = 4.5 - 1

5.5 X = 3.5

X = 3.5 / 5.5

  X = 0.64 M

[ B ] at equilibrium = 1 + X= 1+ 0.64 M = 1.64 M

[ A ] at equilibrium = 1-X = 1- 0.64 M = 0.36 M

Fraction of B moles = moles of B at equilibrium / total moles of A and B

= 1.64 / 1.64+ 0.36

= 1.64/2

= 0.82

= 82 %