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Chapter 17: Problem 6

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
To determine the correct sign for the relationship between \(\Delta G^\circ\) and \(RT \ln(K)\), we need to analyze the relationship between spontaneous reactions and the equilibrium constant. For spontaneous reactions with \(ΔG < 0\), the equilibrium constant \(K > 1\), implying that the natural logarithm of K is positive (\(ln(K) > 0\)). For non-spontaneous reactions with \(ΔG > 0\), the equilibrium constant \(K < 1\), implying that the natural logarithm of K is negative (\(ln(K) < 0\)). Thus, \(\Delta G^\circ\) and \(ln(K)\) have opposite signs, and the correct equation is: \[ \Delta G^\circ = -RT \ln(K) \]

Step by step solution

01

Understanding the meaning of Gibbs free energy change (

) Gibbs free energy change (ΔG) represents the maximum amount of reversible work a chemical reaction can perform. When ΔG is negative, the reaction is spontaneous and proceeds in the forward direction. When ΔG is positive, the reaction is non-spontaneous and proceeds in the reverse direction. When ΔG is zero, the reaction is at equilibrium.
02

Understanding the meaning of the equilibrium constant K (

) The equilibrium constant (K) is the ratio of products to reactants in a chemical reaction at equilibrium. When K > 1, there is a larger concentration of products than reactants, implying that the reaction favors the forward direction. When K < 1, there is a larger concentration of reactants than products, implying that the reaction favors the reverse direction. When K = 1, the concentration of products is equal to the concentration of reactants, and the reaction is at equilibrium.
03

Relate ΔG to K using spontaneity and equilibrium position (

) Now we need to connect Gibbs free energy change with the equilibrium constant. - When the reaction is spontaneous (ΔG < 0), it favors the formation of products, so K > 1. - When the reaction is non-spontaneous (ΔG > 0), it favors the formation of reactants, so K < 1. - When the reaction is at equilibrium (ΔG = 0), the concentrations of products and reactants are equal, so K = 1.
04

Determine the correct sign in the relationship between ΔG and RT ln(K) (

) Based on the relationship between ΔG and K from Step 3, we can conclude the following: - If K > 1, ΔG should be negative. This means that in the equation ΔG = RT ln(K), the natural logarithm of K must also be positive. Hence, K > 1 implies ln(K) > 0. - If K < 1, ΔG should be positive. This means that in the equation ΔG = RT ln(K), the natural logarithm of K must also be negative. Hence, K < 1 implies ln(K) < 0. - If K = 1, ΔG should be zero. This means that ln(K) should also be zero when K = 1. Considering these relationships, ΔG and ln(K) must have opposite signs. Therefore, the correct equation is: \[ \Delta G^\circ = -RT \ln(K) \]

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Most popular questions from this chapter

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\) a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process $\mathrm{S}_{\text { rhombic}}(s)\longrightarrow \mathrm{S}_{\text { monoclinic }}(s)$ b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

For rubidium \(\Delta H_{\mathrm{vap}}^{\circ}=69.0 \mathrm{kJ} / \mathrm{mol}\) \(686^{\circ} \mathrm{C},\) its boiling point. Calculate $\Delta S^{\circ}, q, w,\( and \)\Delta E$ for the vaporization of 1.00 mole of rubidium at \(686^{\circ} \mathrm{C}\) and 1.00 atm pressure.

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. $2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}_{\text { rhombic}}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$ b. $2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$ c. $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$

Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta G^{\circ}=-92.50 \mathrm{kJ}\) Which of the following statements is (are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K\) is greater than 1.00 .

When most biological enzymes are heated, they lose their catalytic activity. This process is called denaturing. The change original enzyme \(\rightarrow\) 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.
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