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Chapter 4: Problem 7

Calculate \(K_{\mathrm{eq}}\) for a reaction with \(\Delta G^{\circ}=-17.1 \mathrm{~kJ} / \mathrm{mol}(-4.09 \mathrm{kcal} / \mathrm{mol})\) at \(328 \mathrm{~K}\). Compare this value to the \(1 \times 10^{3}\) seen at \(298 \mathrm{~K}\).

Short Answer

Expert verified
In this exercise, we calculated the equilibrium constant (K_eq) for a reaction at 328 K with a given standard Gibbs free energy change (∆G°) of -17.1 kJ/mol, using the relationship ∆G° = -RT ln(K_eq). We found K_eq at 328 K to be approximately 6.51 x 10^3. Comparing this value to the given K_eq at 298 K (1 x 10^3), we observed that the equilibrium constant increased approximately 6.5 fold as the temperature increased from 298 K to 328 K.

Step by step solution

01

Convert the given values to appropriate units

The given standard Gibbs free energy change is -17.1 kJ/mol. For calculation purposes, we will need to convert this value to the same energy unit as the gas constant R, which is in J/mol. To convert from kJ/mol to J/mol, simply multiply by 1,000: $$-17.1 \ (kJ/mol) = -17,100 \ (J/mol)$$
02

Rearrange the equation to solve for \(K_{eq}\)

Now we need to rearrange the equation to solve for the equilibrium constant: $$K_{eq} = \mathrm{e}^{\frac{-\Delta G^{\circ}}{RT}}$$
03

Calculate the equilibrium constant at 328 K

Now that we have the equation and the units are consistent, we can plug in the given values for \(\Delta G^{\circ}\), \(R\), and \(T\) to calculate \(K_{eq}\): $$K_{eq} = \mathrm{e}^{\frac{-(-17,100 \ J/mol)}{(8.31 \ J/(mol \cdot K)) \cdot (328 \ K)}}$$ $$K_{eq} \approx 6.51 \times 10^{3}$$
04

Compare the equilibrium constant at 328 K to the given value at 298 K

The given equilibrium constant at 298 K is \(1 \times 10^3\). To compare this value to the calculated equilibrium constant at 328 K, we can observe that the equilibrium constant increases as the temperature increases: $$K_{eq}(298 \ K) = 1 \times 10^3$$ $$K_{eq}(328 \ K) \approx 6.51 \times 10^3$$ The equilibrium constant has increased approximately 6.5 fold as the temperature increased from 298 K to 328 K. This comparison helps us in understanding how the equilibrium constant varies with temperature for this particular reaction.

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Most popular questions from this chapter

Following is a structural formula for the tert-butyl cation. (We discuss the formation, stability, and reactions of cations such as this one in Chapter 6 .) C[C+](C)C tert-Butyl cation (a carbocation) (a) Predict all \(\mathrm{C}-\mathrm{C}-\mathrm{C}\) bond angles in this cation. (b) What is the hybridization of the carbon bearing the positive charge? (c) Write a balanced equation to show its reaction as a Lewis acid with water. (d) Write a balanced equation to show its reaction as a Brønsted-Lowry acid with water.

If the \(\Delta G^{\circ}\) for a reaction is \(-4.5 \mathrm{kcal} / \mathrm{mol}\) at \(298 \mathrm{~K}\), what is the \(K_{\text {eq }}\) for this reaction? What is the change in entropy of this reaction if \(\Delta H^{\circ}=-3.2 \mathrm{kcal} / \mathrm{mol}\) ?

As we shall see in Chapter 19 , hydrogens on a carbon adjacent to a carbonyl group are far more acidic than those not adjacent to a carbonyl group. The anion derived from acetone, for example, is more stable than is the anion derived from ethane. Account for the greater stability of the anion from acetone. $$ \begin{array}{cc} \stackrel{O}{\|} & \\ \mathrm{CH}_{3} \mathrm{CCH}_{2}-\mathrm{H} & \mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{H} \\ \text { Acetone } & \text { Ethane } \\ \mathrm{p} K_{\mathrm{a}} 22 & \mathrm{p} K_{\mathrm{a}} 51 \end{array} $$

Glutamic acid is another of the amino acids found in proteins (Chapter 27). Glutamic acid has two carboxyl groups, one with \(\mathrm{p} K_{\mathrm{a}} 2.10\) and the other with \(\mathrm{p} K_{\mathrm{a}} 4.07\). [NH3+]C(CCC(=O)O)C(=O)O Glutamic acid (a) Which carboxyl group has which \(p K_{a}\) ? (b) Account for the fact that one carboxyl group is a considerably stronger acid than the other.

Predict the position of equilibrium, and calculate the equilibrium constant, \(K_{\text {cq, }}\), for each acid-base reaction. (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\) \(\begin{array}{ccc}\text { Methylamine Aceticacid } & \begin{array}{c}\text { Methylammonium } \\ \text { ion }\end{array} & \begin{array}{c}\text { Acetate } \\ \text { ion }\end{array}\end{array}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{NH}_{2}{ }^{-}\) Ethoxide ion Ammonia Ethanol Amide ion
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