Question

In a unimolecular reaction with twice 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.

Answer 1

Answer : The value of $K_{eq}$ is, 0.5 and $\Delta G$= positive.

Explanation :

The unimolecular reaction is:

$A\rightarrow B$

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

$A=2B$      .........(1)

As we know that:

$K_{eq}=\frac{B}{A}$     ...........(2)

Now substitute equation 1 in 2, we get:

$K_{eq}=\frac{\frac{A}{2}}{A}$

$K_{eq}=0.5$

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 = 0.5

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

$\Delta G^o=-(8.314J/mol.K)\times (298K)\ln (0.5)$

$\Delta G^o=1717.32J/mol$

Thus, the value of $\Delta G^o$ at 298 K is, 1717.32 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.