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Chapter 15: Problem 39

Two different proteins \(X\) and \(Y\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\) . The proteins bind in a \(1 : 1\) ratio to form \(X Y . A\) 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 \(\mathrm{mM}\) of free Y remain. What is \(K_{c}\) for the reaction?

Short Answer

Expert verified
The equilibrium constant, \(K_c\), for the reaction \(X + Y \rightleftharpoons XY\) can be determined using the given initial and equilibrium concentrations of X and Y. From the ICE table, we find that the change in concentration (x) is 0.80 mM and the equilibrium concentration of XY is also 0.80 mM. Using the formula \(K_c = \frac{[XY]}{[X][Y]}\) and substituting the equilibrium concentrations, we find that \(K_c = \frac{0.80}{(0.20)(0.20)} = 20\). Therefore, the equilibrium constant for this reaction is 20.

Step by step solution

01

Write the balanced chemical equation

Firstly, we need to write the balanced chemical equation for the reaction. Since the proteins X and Y bind in a 1:1 ratio to form XY, we can write the chemical equation as: \(X + Y \rightleftharpoons XY\)
02

Set up the ICE table

To find the equilibrium constant (\(K_c\)), we need to know the equilibrium concentrations of all the species involved in the reaction. We are given the initial and equilibrium concentrations of X and Y. We can set up an Initial, Change, and Equilibrium (ICE) table to represent the changes in concentrations during the reaction. | | X | Y | XY | |--------|------|------|------| | Initial| 1.00 | 1.00 | 0 | | Change | -x | -x | +x | | Equilibrium | 0.20 | 0.20 | ? |
03

Determine the change in concentration

From the ICE table, we can determine the change in concentration (x) for each species using the given equilibrium concentrations of X and Y: \(x = 1.00 - 0.20 = 0.80\)
04

Find the equilibrium concentration of XY

Now that we have found the change in concentration (x), we can determine the equilibrium concentration of XY: Initial concentration of XY + Change in concentration = Equilibrium concentration of XY \(0 + 0.80 = 0.80\,\text{mM}\)
05

Calculate the equilibrium constant, \(K_c\)

The equilibrium constant, \(K_c\), can be calculated using the equilibrium concentrations of the species in the chemical equation: \(K_c = \frac{[XY]}{[X][Y]}\) Substitute the equilibrium concentrations: \(K_c = \frac{0.80}{(0.20)(0.20)}\) Calculate the value of \(K_c\): \(K_c = \frac{0.80}{0.04} = 20\) Thus, the equilibrium constant for the reaction is 20.

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Most popular questions from this chapter

(a) If \(Q_{c} < K_{O}\) in which direction will a reaction proceed in order to reach equilibrium? (b) What condition must be satisfied so that \(Q_{c}=K_{c} ?\)

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

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 \(\mathrm{H}_{2} \mathrm{SO}_{4} ) .\) A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v) & \rightleftharpoons \\ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\mathrm{solv}) &+\mathrm{H}_{2} \mathrm{O}(\text {solv}) \end{aligned}$$ where \(^{a}(s o l v)^{\prime \prime}\) 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\(M\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and \(\mathrm{HI}\) at \(458^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{mol} \mathrm{H}_{2}, 0.112 \mathrm{mol} \mathrm{I}_{2},\) and 0.775 \(\mathrm{mol}\) HI in a 5.00 -L. vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 \(\mathrm{mol}\) of \(\mathrm{HI}\) ?

NiO is to be reduced to nickel metal in an industrial process by use of the reaction $$\mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g)$$ At \(1600 \mathrm{K},\) the equilibrium constant for the reaction is \(K_{p}=6.0 \times 10^{2} .\) If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?
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