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Chapter 15: Problem 39

Two different proteins \(\mathrm{X}\) and \(\mathrm{Y}\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\). The proteins bind in a 1: 1 ratio to form XY. A solution that is initially \(1.00 \mathrm{mM}\) in each protein is allowed to reach equilibrium. At equilibrium, \(0.20 \mathrm{mM}\) of free \(\mathrm{X}\) and \(0.20 \mathrm{~m} M\) of free Y remain. What is \(K_{c}\) for the reaction?

Short Answer

Expert verified
The equilibrium constant \(K_c\) for the reaction between proteins X and Y, which bind in a 1:1 ratio to form XY, is 20. This was calculated using the given initial and equilibrium concentrations and the formula for the equilibrium constant.

Step by step solution

01

Write down the reaction

The given reaction is: \[X + Y \rightleftharpoons XY\] The stoichiometric coefficients for X, Y, and XY are all 1.
02

Calculate the equilibrium concentrations

Use the initial concentrations and equilibrium concentrations of X and Y to determine the equilibrium concentration of XY. Initial concentrations: [X] = [Y] = 1.00 mM Equilibrium concentrations: [X]\(_{eq}\) = [Y]\(_{eq}\) = 0.20 mM The equilibrium concentration of XY can be calculated by comparing the difference in initial and equilibrium concentrations of X (or Y): [XY]\(_{eq}\) = 1.00 mM - 0.20 mM = 0.80 mM.
03

Write the formula for \(K_c\)

For this reaction, the formula for the equilibrium constant \(K_c\) can be written as: \[K_c = \frac{[\mathrm{XY}]_{eq}}{[\mathrm{X}]_{eq} [\mathrm{Y}]_{eq}}\]
04

Calculate \(K_c\)

Now we can use the equilibrium concentrations calculated in Step 2 to find the value of \(K_c\): \[K_c = \frac{0.80 \text{ mM}}{(0.20 \text{ mM})(0.20 \text{ mM})} = \frac{0.80}{0.04} = 20\] Therefore, the equilibrium constant \(K_c\) for the reaction is 20.

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Most popular questions from this chapter

As shown in Table \(15.2,\) the equilibrium constant for the reaction $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\( is \)K_{p}=4.23 \times 10^{-7}\( at \)300^{\circ} \mathrm{C}\(. Pure \)\mathrm{NH}_{3}$ is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are $1.05 \mathrm{~g} \mathrm{NH}_{3}$ in the equilibrium mixture. (a) What are the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$. A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up $15.0 \mathrm{~L}\( of a solution that is initially \)0.275 \mathrm{M}$ in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction $4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\( in a fetus, compared to \)K_{c}$ for the same reaction in an adult.

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15 ), \(K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(g) .\) A 7.5-L gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g)\), which is allowed to equilibrate at 450 \(\mathrm{K} .\) At equilibrium the partial pressures of the three gases are $P_{\mathrm{PCl}_{3}}=12.56 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=15.91 \mathrm{kPa},\( and \)P_{\mathrm{PCl}_{5}}=131.7 \mathrm{kPa}$ (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at \(450 \mathrm{~K}\).
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