Question

In a unimolecular reaction with three times as much starting material as product at equilibrium, what is the value of Keq? Is ΔG o positive or negative? Enter Keq as a decimal. Be sure to answer all parts. Keq =

Answer 1

Answer : The value of
`K_(eq)` is, 0.33 and
`\Delta G`= positive.

Explanation :

The unimolecular reaction is:

`A\rightarrow B`

In unimolecular reaction, the starting material is 2 times to the product.

`A=3B` .........(1)

As we know that:

`K_(eq)=(B)/(A)` ...........(2)

Now substitute equation 1 in 2, we get:

`K_(eq)=((A)/(3))/(A)`

`K_(eq)=0.33`

Now we have to calculate the value of
`\Delta G^o` at 298 K.

`\Delta G^o=-RT\ln K_(eq)`

where,

`\Delta G^o` = standard Gibbs free energy = ?

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K

`K_(eq)` = equilibrium constant = 0.33

Now put all the given values on the above formula, we get:

`\Delta G^o=-(8.314J/mol.K)* (298K)\ln (0.33)`

`\Delta G^o=2746.8J/mol`

Thus, the value of
`\Delta G^o` at 298 K is, 2746.8 J/mol

As we know that:

`\Delta G`= +ve, reaction is non spontaneous

`\Delta G`= -ve, reaction is spontaneous

`\Delta G`= 0, reaction is in equilibrium

Thus, the
`\Delta G`= +ve. So, the reaction is non spontaneous.