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
Recall that a negative ∆G° indicates a spontaneous reaction that favors products (K > 1), while a positive ∆G° represents a non-spontaneous reaction that favors reactants (K < 1). Testing the possible expressions, we can conclude that the correct relationship is ∆G° = -RTln(K), as it establishes the appropriate connection between ∆G° and K values.

Step by step solution

01

Identify the meaning of ∆G° and K

First, let's recall that ∆G° is the standard Gibbs free energy change which indicates whether a reaction is thermodynamically favorable or unfavorable. A negative ∆G° implies a spontaneous reaction, and a positive ∆G° means a non-spontaneous reaction. K is the equilibrium constant, which represents the ratio of the concentrations (or activities) of the products to the reactants at equilibrium. A larger K value indicates that the reaction favors the formation of products, while a smaller K value means that the reactants are favored. When K is greater than 1, the reaction favors products, and when K is less than 1, the reaction favors reactants.
02

Analyze the connection between ∆G° and K

Now let's consider the possible cases to understand the relationship between ∆G° and K: 1. If ∆G° is negative (favorable reaction), the reaction will proceed spontaneously, meaning the products are favored over the reactants. In this case, we expect K to be greater than 1. 2. If ∆G° is positive (unfavorable reaction), the reaction is not spontaneous, indicating that the reactants are favored over the products. In this case, we expect K to be less than 1.
03

Determine the appropriate sign for the expression

We have two possible expressions, either ∆G° = RTln(K) or ∆G° = -RTln(K). We will consider both cases to see which one matches the relationship between ∆G° and K deduced in Step 2. Case 1: ∆G° = RTln(K) If ∆G° is negative, then ln(K) must also be negative, which implies K < 1. However, this goes against our understanding of the relationship between ∆G° and K, as a negative ∆G° should favor products (K > 1). Case 2: ∆G° = -RTln(K) If ∆G° is negative, then ln(K) must be positive, which implies K > 1. This matches our understanding of the relationship between ∆G° and K as deduced in Step 2. Based on this analysis, we conclude that the correct sign in the expression is negative, and the correct relationship is: ∆G° = -RTln(K)

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Most popular questions from this chapter

The equilibrium constant \(K\) for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 71 .

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix 4, predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature dependent. Predict the sign of \(\Delta S_{\text {sarr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) at \(298 \mathrm{~K}\). Using these values and data in Appendix 4, calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the preceding reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\).

For the reaction at \(298 \mathrm{~K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ}\) and \(-176.6 \mathrm{~J} / \mathrm{K}, \mathrm{re}-\) spectively. What is the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) ? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G\) negative above or below this temperature?

When most biologic enzymes are heated, they lose their catalytic activity. The change Original enzyme \(\longrightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

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