Question

Question: Under base-catalyzed conditions, two molecules of acetone can condense to form diacetone alcohol. At room temperature${25}^{0}c$, aboutof 5%the acetone is converted to diacetone alcohol. Determine the value of$∆G^0$for this reaction.

Short answer

Answer

The value of$∆G^0$ is found to be 7.2KJ/mol.

Step-by-step solution

Equilibrium constant (keq)

The equilibrium constant (k_eq) of the reaction governs the equilibrium concentration of the reactants and products.For a general reaction of the type, $aA+bB\leftrightharpoons cC+dD$) can be written as follows:

$$\begin{gathered} K_{eq}=\left[\frac{products}{reactents}\right] \\ =\left[\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}\right] \end{gathered}$$

Position of equilibrium

The position of equilibrium can be predicted from the value of the equilibrium constant (k_eq) . The forward reaction is favored if(k_eq) > 1 , and the backward reaction is favored if (k_eq)< 1.

Explanation

Since 5% of the acetone is converted to diacetone alcohol, we can write it as follows:[diacentone alcohol] = 5% = 0.05

[acetone]= 95%= 0.95

Now,

$$\begin{gathered} k_{eq}=\left[\frac{products}{reactents}\right] \\ \Rightarrow k_{eq}=\left[\frac{dieacetonealcohol}{{\left[acetone\right]}^2}\right] \\ \Rightarrow k_{eq}=\left[\frac{0.05}{{[0.95]}^2}\right] \\ \Rightarrow k_{eq}=\frac{0.05}{0.9025} \\ \Rightarrow k_{eq}=0.055 \end{gathered}$$

$$\begin{gathered} Theexpressionfors\tan dardGibb’sfreeenergyis∆G^0=-RT\ln k_{eq}. \\ Thisexpressioncanalsobewrittenas∆G^0=-2.303RT\log k_{eg}.Asperthegivendata, \\ T={25}^{0}C=(25+273)K=298K \\ R=8.314J/mol(universalgascons\tan t) \\ {k}_{eq}=0.055\left(calculated\right) \\ Again, \\ Findthevalueof∆G^{0}asfollows: \\ ∆G^0=-2.303RT\log k_{eg} \\ =-2.303\times 8.314J/K.mol\times 298k\times T\log (0,055) \\ =-5705.85\times -1.26 \\ =7189.371J/Mol \\ =7.2KJ/mol \\ Thevalueof∆G^{0}isfoundtobe7.2KJ/mol. \end{gathered}$$