Question

a. Which value corresponds to a negative value of $\mathbf{∆}\mathbf{G}\mathbf{°}\mathbf{:}\mathbf{}{\mathbf{K}}_{\mathbf{eq}}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$or ${\mathbf{K}}_{\mathbf{eq}}\mathbf{=}{\mathbf{10}}^{\mathbf{2}}$?

b. In a unimolecular reaction with five times as much starting material as product at equilibrium, what is the value of K_eq? Is $\mathbf{∆}\mathbf{G}\mathbf{°}$positive or negative?

c. Which value corresponds to a larger ${\mathbf{K}}_{\mathbf{eq}}$: $\mathbf{∆}\mathbf{G}\mathbf{°}\mathbf{=}\mathbf{-}\mathbf{8}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$or $\mathbf{∆}\mathbf{G}\mathbf{°}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{kJ}\mathbf{/}\mathbf{mol}$?

Short answer

Answer

1. For a negative $∆\mathrm{G}°$, the value of ${\mathrm{K}}_{\mathrm{eq}}$is ${10}^2$.
2. ${\mathrm{K}}_{\mathrm{eq}}<1.$So, the reactants are favored at equilibrium. Hence, the value of is positive.
3. A negative $∆\mathrm{G}°$has a value of ${\mathrm{K}}_{\mathrm{eq}}$greater than one. A positive $∆\mathrm{G}°$has the value of ${\mathrm{K}}_{\mathrm{eq}}$lesser than one.

Hence, $∆\mathrm{G}°=-8\mathrm{kJ}/\mathrm{mol}$corresponds to a larger ${\mathrm{K}}_{\mathrm{eq}}$.

Step-by-step solution

Step-by-Step SolutionStep 1: Keq and ∆G°

For the spontaneous reactions in the direction of products, the value of $\mathbf{∆}\mathbf{G}\mathbf{°}$is negative. The value of ${\mathbf{K}}_{\mathbf{eq}}$favoring the products should be greater than one and vice-versa.

Equilibrium Constant

${\mathrm{K}}_{\mathrm{eq}}$is less than one when the equilibrium favors the reactants; however, the value of ${\mathrm{K}}_{\mathrm{eq}}$is greater than one when the equilibrium favors the products.

Explanation

a. A negative value of $∆\mathrm{G}°$means that the reaction is spontaneous in the direction of products. If the reaction favors the products, the ${\mathrm{K}}_{\mathrm{eq}}>1$. Hence, for a negative $∆\mathrm{G}°$, the value of ${\mathrm{K}}_{\mathrm{eq}}$is ${10}^2$.

b. Suppose the concentration of the products at equilibrium is 1 unit. Hence, the concentration of reactants

at equilibrium is 5 units.

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{eq}}& =& \frac{\left[\mathrm{Products}\right]}{\left[\mathrm{Reactants}\right]}\\ & =& \frac{\left[1\right]}{\left[5\right]}\\ & =& 0.2\end{array}$

Hence, ${\mathrm{K}}_{\mathrm{eq}}<1$. So, the reactants are favored at equilibrium. Hence, the value of $∆\mathrm{G}°$is positive.

c. A negative $∆\mathrm{G}°$has a value of ${\mathrm{K}}_{\mathrm{eq}}$greater than one. A positive $∆\mathrm{G}°$has the value of ${\mathrm{K}}_{\mathrm{eq}}$lesser than one. Hence, $∆\mathrm{G}°=-8\mathrm{kJ}/\mathrm{mol}$corresponds to a larger ${\mathrm{K}}_{\mathrm{eq}}$.