Question

Consider the following diagram of free energy (G) versus fraction of A reacted in terms of moles for the reaction $\mathbf{2}\mathit{A}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathit{B}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

Before any A has reacted, ${\mathit{P}}_{\mathbf{A}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathit{a}\mathit{t}\mathit{m}\mathit{a}\mathit{n}\mathit{d}{\mathit{P}}_{\mathbf{B}}\mathbf{=}\mathbf{0}\mathbf{.}$determine the sign of$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$and the value of K for this reaction.

Short answer

The sign of${\text{ΔG}}^{\text{o}}$ is positive.

The value of $\text{K\hspace{0.17em}is}\frac{1}{8}{\text{atm}}^{\text{-1}}$before any A has reacted

Step-by-step solution

Definition of standard free energy

The amount of energy released in the conversion of reactants to products under standard conditions is defined as standard free energy.

Find equilibrium pressure

The reaction is $\text{2A(g)}\to \text{B(g)}$

The pressure at equilibrium point is${\text{P}}_{\text{A}}^{\text{eq}}=2.00\text{\hspace{0.17em}atm}$

The ratio between $\text{A\hspace{0.17em}and\hspace{0.17em}B\hspace{0.17em}is\hspace{0.17em}2:1}$

$\begin{array}{l}{\text{P}}_{\text{B}}^{\text{eq}}=\frac{1}{2}\left(\frac{1}{3}\right)\times \left(3.00\text{\hspace{0.17em}atm}\right)\\ {\text{P}}_{\text{B}}^{\text{eq}}=0.5\text{\hspace{0.17em}atm}\end{array}$

The equilibrium constant is$\text{K}=\frac{1}{8}{\text{atm}}^{\text{-1}}$

Sign of${\text{ΔG}}^{\text{o}}=\text{RT}\ln \left(8\right)$

So, the sign of ${\text{ΔG}}^{\text{o}}$is positive and value of $\text{K\hspace{0.17em}is}\frac{1}{8}{\text{atm}}^{\text{-1}}$before any A has reacted.