Question

Semi-Challenge Question*

Answer 1

Since everyone is posting "challenge" questions in other sections, I decided to start posting some "challenge" chemistry questions from previous exams or review guides. So here goes:

\(X \leftrightarrow Y\)
the above reaction shows the instantaneous free energy \(\Delta G\) of the reaction and is defined as: \(\Delta G=RT~ln \frac{Q}{K}\) where \(R\) is the gas constant: \(8.314 \frac{J}{mol~K}\), \(T\) is \(temperature~ (K)\), \(Q\) is the reaction quotient, & \(K\) is your equilibrium constant. Under what conditions does \(\Delta G = \Delta G^{\circ}\), which is the \(standard~reaction~free~energy\).

Answer 2

\(\boxed{\mathsf{\Delta _ r G = \Delta _ r G^{0} + RT \ln Q }}\) \(----- \mathsf{equation} 1\)

\(\textbf{Where : Q is the reaction quotient.}\\

\textbf{When equilibrium is attained, \(\Delta G = 0 \) and thus : }\)

\(\mathsf{0 = \Delta _ r G^{0} + RT \ln K} \) \(\textbf{(As Q becomes equal to K (equilibrium constant)) }\)

\(\mathsf{-RT \ln K = \Delta _r G^{0}} \)

\(\mathsf{Therefore,} \space \textbf{equation 1 becomes :} \)

\(\mathsf{\Delta _ r G = -RT \ln K + RT \ln Q }\)

\(\mathsf{\Delta_r G = RT(\ln Q - \ln K)}\)

\(\mathsf{\Delta _ rG = RT\ln \cfrac{Q}{K}}\)

Writing equation 1 again,

\(\boxed{\mathsf{\Delta _ r G = \Delta _ r G^{0} + RT \ln Q }}\)

Using this equation :
\(\mathsf{For } \space \mathsf{\Delta_r G }=\mathsf{\Delta_r G^{0}} \space \mathsf{Q = 1 }\space \mathsf{or }\space \mathsf{T = 0}\space \mathsf{ Kelvin} \)

Answer 3

@abb0t m i correct?