You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

Short Answer

Expert verified
By analyzing the meanings of \(\Delta G^{\circ}\) and \(K\), we can deduce that a more negative \(\Delta G^{\circ}\) corresponds to a larger value of \(K\), and a more positive \(\Delta G^{\circ}\) corresponds to a smaller value of \(K\). The correct relationship between \(\Delta G^{\circ}\) and \(R T \ln(K)\) that aligns with these observations is \(\Delta G^{\circ} = -R T \ln(K)\).

Step by step solution

01

Understanding the definitions and meanings of \(\Delta G^{\circ}\) and \(K\)

\(\Delta G^{\circ}\) is the standard change in Gibbs free energy, which is a measure of the maximum reversible work that can be done by a system at constant temperature and pressure. A negative \(\Delta G^{\circ}\) indicates a spontaneous reaction, while a positive \(\Delta G^{\circ}\) implies a non-spontaneous reaction. \(K\) is the equilibrium constant, representing the ratio of the concentrations of products to reactants at equilibrium. If \(K > 1\), the reaction favors the formation of products, while if \(K < 1\), the reaction favors the formation of reactants.
02

Relating the magnitude of \(\Delta G^{\circ}\) to the value of \(K\)

We can reason that if a reaction has a negative \(\Delta G^{\circ}\) (spontaneous), it should favor the formation of products, so \(K > 1\) in this case. Conversely, if a reaction has a positive \(\Delta G^{\circ}\) (non-spontaneous), it should favor the formation of reactants, so \(K < 1\) in this situation. From these observations, we can conclude that a more negative \(\Delta G^{\circ}\) corresponds to a larger value of \(K\), and a more positive \(\Delta G^{\circ}\) corresponds to a smaller value of \(K\)
03

Determining the sign in the relationship between \(\Delta G^{\circ}\) and \(R T \ln(K)\)

Now, let's consider the relationship between \(\Delta G^{\circ}\) and \(R T \ln(K)\). Based on Step 2, if the value of \(K\) increases, the value of \(\Delta G^{\circ}\) should become more negative. If the relationship were given as \(\Delta G^{\circ} = R T \ln(K)\), then as \(K > 1\), we would have a positive \(\ln(K)\). In this case, the product of \(R T \ln(K)\) would also be positive, which contradicts our observation that a larger \(K\) should correspond to a more negative \(\Delta G^{\circ}\). However, if the relationship were given as \(\Delta G^{\circ} = -R T \ln(K)\), then as \(K > 1\), we would again have a positive \(\ln(K)\), but this time the negative sign in the relationship would make \(\Delta G^{\circ}\) more negative, which is consistent with our observation in Step 2. Therefore, the correct relationship between \(\Delta G^{\circ}\) and \(R T \ln(K)\) is: \[\Delta G^{\circ} = -R T \ln(K)\]

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Most popular questions from this chapter

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ}\), for a reaction at \(25^{\circ} \mathrm{C}\). How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C} ?\) What assumptions are made?

Predict the sign of \(\Delta S_{\text {surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}_{2}(s)\)

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}}\), where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{\mathrm{a}} / R T}\right) .\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

Which of the following involve an increase in the entropy of the system? a. melting of a solid b. sublimation c. freezing d. mixing e. separation f. boiling

The value of \(\Delta G^{\circ}\) for the reaction $$ 2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-5490 . \mathrm{kJ} .\) Use this value and data from Appendix 4 to calculate the standard free energy of formation for \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\).

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