Answer : The value of [tex]K_{eq}[/tex] is, 0.33 and [tex]\Delta G[/tex]= positive.
Explanation :
The unimolecular reaction is:
[tex]A\rightarrow B[/tex]
In unimolecular reaction, the starting material is 2 times to the product.
[tex]A=3B[/tex] .........(1)
As we know that:
[tex]K_{eq}=\frac{B}{A}[/tex] ...........(2)
Now substitute equation 1 in 2, we get:
[tex]K_{eq}=\frac{\frac{A}{3}}{A}[/tex]
[tex]K_{eq}=0.33[/tex]
Now we have to calculate the value of [tex]\Delta G^o[/tex] at 298 K.
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = ?
R = gas constant = 8.314 J/mol.K
T = temperature = 298 K
[tex]K_{eq}[/tex] = equilibrium constant = 0.33
Now put all the given values on the above formula, we get:
[tex]\Delta G^o=-(8.314J/mol.K)\times (298K)\ln (0.33)[/tex]
[tex]\Delta G^o=2746.8J/mol[/tex]
Thus, the value of [tex]\Delta G^o[/tex] at 298 K is, 2746.8 J/mol
As we know that:
[tex]\Delta G[/tex]= +ve, reaction is non spontaneous
[tex]\Delta G[/tex]= -ve, reaction is spontaneous
[tex]\Delta G[/tex]= 0, reaction is in equilibrium
Thus, the [tex]\Delta G[/tex]= +ve. So, the reaction is non spontaneous.
