Chapter 15, Problem 15.156QP

At present, the World Anti-Doping Agency has no way to detect the use of hypoxic sleeping tents, other than inspection of an athlete’ s home. Imagine, however, that a biochemical analysis company is developing a way to determine whether an elevated red blood cell count in an athlete’s blood is the result of such a practice. A key substance in the detection process is the proprietary compound OD17X, which is produced by the combination of two other proprietary compounds. OD1A and OF2A The aqueous reaction is represented by the equation

$\begin{array}{l}\text{OD1A}\left(aq\right)+\text{OF2A}\left(aq\right)\rightleftarrows \text{OD17X}\left(aq\right)\\ \Delta H=\text{29 kJ/mol}\end{array}$

(a) Write the equilibrium expression for this reaction. (b) Given the equilibrium concentrations at room temperature of [OD1 A) = 2.12 M, [OF2A] = 1.56 M, and (OD17X) = 1.01 × 10⁻⁴ M, calculate the value of the equilibrium constant (K_c) at room temperature. (c) The OD17X produced is precipitated with another proprietary substance according to the equation

$\text{OD17X}\left(aq\right)+\text{A771A}\left(aq\right)\rightleftarrows \text{OD17X}-\text{A77}\left(s\right)$

K_c for the precipitation equilibrium is 1.0 × 10⁶ at room temperature. Write the equilibrium expression for the sum of the two reactions and determine the value of the overall equilibrium constant. (d) Determine whether a mixture with [OD1A] = 3.00 M, [OF2A] = 2.50 M, and [OD17X] = 2.7 x 10⁻⁴ M is at equilibrium and, if not, which direction it will have to proceed to achieve equilibrium. (e) Because the production of OD17X is endothermic, it can be enhanced by increasing the temperature. At 250°C, the equilibrium constant for the reaction

$\text{OD1A}\left(aq\right)+\text{OF2A}\left(aq\right)\rightleftarrows \text{OD17X}\left(aq\right)$

is 3.8 x 10². If a synthesis at 250°C begins with 1.0 M of each reactant, what will be the equilibrium concentrations of reactants and products?

Interpretation Introduction

Interpretation:

The different chemical equilibrium terms should be derived given the biological importance equilibrium reactions with supporting from multiple concept of given statements.

Concept Introduction:

Equilibrium constant (Kc): Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Temperature affect in equilibrium: This process chemical shifts changes (or) towards the product or reactant, which can be determined by studying the reaction and deciding whether it is exothermic or endothermic. In other words if reactions are absorption for heat from their surroundings and have a $(+\Delta H)$ positive then many of reactions are evaporating for heat from their surroundings and enthalpy values is $(-\Delta H)$ negative.

Free energy: The change in the standard energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy $(\text{ΔH)}$ of the system minus the change in the respective product of the temperature at the times of entropy of the system, the following equation $\text{G=H-(TS) or (ΔG=ΔH-TΔS)}$.

Standard free energy $({\text{ΔG}}^0)$: The entropy is second law of thermodynamics it indicated for $({\text{ΔG}}^0)$ symbol, the many of chemical reactions cause changes in entropy and it plays on important role in determining in which direction (forward and backward) a chemical reaction spontaneously proceeds.

Heterogeneous equilibrium: This equilibrium reaction does not depend on the amounts of pure solid and liquid present, in other words heterogeneous equilibrium, substances are in different phases.

Partial pressure (Kp):  The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.

Forward Reaction: This type of reaction has involved irreversible, if obtained product cannot be converted back in to respective reactants under the same conditions. Backward Reaction: This type of reaction process involved a reversible, if the products can be converted into a back to reactants.

Reaction quotient: This type of chemical equilibrium reaction proceeds likely to produced, given either the pressure (or) the concentration of the reactants and the products. The value can be compared to the equilibrium constant, to determine the direction of the reaction that is take place. Then reaction quotient (Qc) the indication of Q can be used to determine which direction will shift to reach of chemical equilibrium process.

Answer to Problem 15.156QP

The reactant and product each chemical equilibrium values (a-c) for the given biological dissociation reactions are shown below.

$\begin{array}{l}\text{a)}\text{.}\text{OD1A(aq)}\text{+}\text{OF2A(aq)}\rightleftharpoons \text{OD17X(aq)}-----\text{ΔH=29}\text{kJ/mol}\\ \text{Kc=}\frac{\text{[OD17X]}}{\text{[OD1A][OF2A]}}\\ \text{b)}\text{.}\text{OD17(aq)}\text{+}\text{A771(aq)}\rightleftharpoons \text{OD17X-A77(s)}\\ \text{Kc=}\text{3}{\text{.05×10}}^{\text{-5}}\\ \text{c)}\text{.}\text{Kc=}\frac{\text{1}}{\text{[OD1A][OF2A][A771A]}}\text{=30}\text{.5}\\ \text{d)}\text{.}\text{Qc=}\frac{\text{[OD17X]}}{\text{[OD1A][OF2A]}}\text{=3}{\text{.6×10}}^{\text{-5}}\\ \text{e)}\text{.}\text{Kc=}\frac{\text{[OD17X]}}{\text{[OD1A][OF2A]}}\\ \text{[OD17X]=0}\text{.95}\text{M,}\text{[OD1A][OF2A]=0}\text{.05M}\end{array}$

Explanation of Solution

To find: The equilibrium expression should be written given the biological reactions.

First we derived the equilibrium pressure values of given statement (a).

Given below the reaction (1), use the law of mass actions to write the equilibrium expression has fallowed.

$\begin{array}{l}\text{a)}\text{.}\text{OD1A(aq)}\text{+}\text{OF2A(aq)}\rightleftharpoons \text{OD17X(aq)}-\text{ΔH=29}\text{kJ/mol--------[1]}\\ \text{Kc=}\frac{\text{[OD17X]}}{\text{[OD1A][OF2A]}}\end{array}$

The (Kc) equation is written by multiplying the activities for the species of the products and dividing by the activities of the reactants. If any component in the reaction has a coefficient, indicated above with lower case letters, the concentration is raised to the power of the coefficient.  The compare for the given above equation (1) and should solved above. Furthermore the Kc changes as the reaction progress and is only equal to the equilibrium constant when total system is at same equilibrium.

To find: Calculate the equilibrium constant $(Kc)$ values for given the statement of biological equilibrium reaction (b).

Calculate and analyze the $(Kc)$ values with respective statement (b).

Here the respective reactant and product molar concentration (M) values are substituted given below equation.

$\begin{array}{l}\text{b)}\text{.}\text{OD17(aq)}\text{+}\text{A771(aq)}\rightleftharpoons \text{OD17X-A77(s)}\\ \text{a)}\text{.}\text{OD1A(aq)}\text{+}\text{OF2A(aq)}\rightleftharpoons \text{OD17X(aq)}\\ \text{Kc=}\frac{\text{[OD17X]}}{\text{[OD1A][OF2A]}}\\ \text{Given molar concentration values are }\\ \text{OD1A}=2.12M,\text{OF2A}=1.56M\to \text{Reactants}\\ \text{OD17X}=1.01\times {10}^{-4}\\ \text{The respactive (M) values are substituted above (Kc) equation}\\ \text{Kc=}\frac{(1.01\times {10}^{-4})}{\text{(}2.12\text{)(}1.56\text{)}}=\frac{(1.01\times {10}^{-4})}{3.3072}=3.05\times {10}^{-5}\\ \text{Kc=}\text{3}{\text{.05×10}}^{\text{-5}}\end{array}$

Given above the equation (b), product will not appeared in this expression, because this molecule has solid phase. So we consider the equilibrium reaction (a) in first statement. The first reaction (a) all in the aqueous phase and derived equilibrium equation (Kc) placed above. Further the molar concentration values are substituted this equation ew get the equilibrium constant $\text{Kc=}\text{3}{\text{.05×10}}^{\text{-5}}$ value. This calculation method showed above.

To find:  The equilibrium expression (K) values should be calculate and analyze the given aqueous phase into solid phase (precipitation) reaction and its statement (c).

$\begin{array}{l}\text{a)}\text{.}\text{OD1A(aq)}\text{+}\text{OF2A(aq)}\rightleftharpoons \text{OD17X(aq)}\\ \text{b)}\text{.}\text{OD17(aq)}\text{+}\text{A771(aq)}\rightleftharpoons \text{OD17X-A77(s)}\\ \text{----------------------------------------------------------}\\ \text{Net reaction: OD1A(aq)}\text{+}\text{OF2A(aq)}\rightleftharpoons \text{OD17X(aq)}\\ {\text{Kc}}_{\text{1}}\text{=}\frac{\text{[OD17X]}}{\text{[OD1A][OF2A]}}------[1]\\ {\text{Kc}}_2\text{=}\frac{1}{\text{[}\text{OD17][A771]}}------[2]\\ Heresolidphasemoleculewillnotappearinthis\text{expression}\\ \text{Thaking}\text{the sum of the two reactions it gives,}\\ \text{a)}\text{.}\text{OD1A(aq)}\text{+}\text{OF2A(aq)}\rightleftharpoons \text{OD17X(aq)}\\ \text{b)}\text{.}\text{OD17(aq)}\text{+}\text{A771(aq)}\rightleftharpoons \text{OD17X-A77(s)}\\ \text{----------------------------------------------------------------------}\\ \text{OD1A(aq)}\text{+}\text{OF2A(aq)}\text{+ A771(aq)}\rightleftharpoons \text{OD17X-A11(s)}\\ \text{The total equilibrium (Kc) has showed below}\\ \text{Kc=}\frac{1}{\text{[}\text{OD17][OF2A][A771]}}------[3]\\ GivenKc\text{and}{\text{Kc}}_{\text{1}}valuesare\text{substituted}\text{equation}(3)\\ \text{Kc=}\frac{1}{(\text{3}{\text{.05×10}}^{\text{-5}})(1.0\times {10}^6)}(or)\\ \text{Kc=}(\text{3}{\text{.05×10}}^{\text{-5}})(1.0\times {10}^6)\\ \text{Kc=}\text{30}\text{.5}\end{array}$

Given the biological equilibrium reaction the equal moles of aqueous phase molecule converted into a solid (OD17X-A77(s) product molecule hence this equilibrium converted into a solid molecule. So we consider the reaction (a) and (b) with respective statements we get the net reaction. The overall equilibrium constants are the produce of the individual constants. Further the derived equilibrium constant values are showed above.

To identify: The equilibrium directions should be identified given the equilibrium reactions with statement (d).

The reactant and product concentrations provided to calculate (Qc) values and then compare with reaction quintet (Qc) with equilibrium constant values (Kc).

$\begin{array}{l}\text{OD1A(aq)}\text{+}\text{OF2A(aq)}\rightleftharpoons \text{OD17X(aq)}\\ \text{Qc=}\frac{\text{[OD17X]}}{\text{[OD1A][OF2A]}}\\ \text{Given the molar concentration values are}\\ \text{[OD1A]=3}\text{.00}\text{M,}\text{[OF2A]=}\text{2}\text{.50}\text{M}\text{and}\text{[OD17X]=2}\text{.7}\times {\text{10}}^{\text{-4}}\\ \text{Respactive}\text{values}\text{are}\text{substituted}\text{above}\text{(Qc)}equation\\ \text{Qc=}\frac{(2.7\times {10}^{-4}M)}{(3.00M)(2.50M)}=\frac{(2.7\times {10}^{-4}M)}{(7.5M)}=3.6\times {10}^{-5}\end{array}$

The obtained values od (Qc) is greater than (Kc). Therefore, the reaction is not at equilibrium and must proceeds to the left to established equilibrium.

To find: Calculate the each concentration values for given the equilibrium constant (Kc) of biological equilibrium reaction.

Calculate and analyze the respective concentration values at $250{}^{\text{0}}\text{C}$.

First we constant an equilibrium table to determine the equilibrium concentration of each species in terms of unknown (x) to solve for (x) and use it to calculate the equilibrium molar concentrations.

Next we starting concentrations that we know into the equilibrium table fallowed below

$\begin{array}{l}\text{OD1A + }\text{OF2A}\rightleftharpoons \text{OD17X}\\ \text{Initial (M): }1.01.00\\ \text{Change (M): }\text{-x}+x+x\\ ----------------------------\\ \text{Eqilibrium (M):}(1.0-x)(1.0-x)x\\ \text{Substitute}\text{into}\text{the }\text{equilibrium}\text{expression}\text{to}\text{solve}\text{for}\text{(x)}\text{.}\\ \text{Kc=}\frac{\text{[OD17X]}}{\text{[OD1A][OF2A]}}----[1]\\ \text{The}\text{starting}\text{equilbrium}\text{concentration}\text{values}\text{are}\\ \text{OD17X}=1.0-x,\text{OF2A}=1.0-x\\ \text{The}\text{given equilibrium constant}\text{of}(Kc)\text{=}3.8\times {10}^2\\ \text{The respactive values are substituted equation (1)}\\ 3.8\times {10}^2=\frac{x}{(1-x)(1-x)}=\frac{x}{1-2x+x^2}-------[2]\\ Rearrangingequation(2)\\ 3.8\times {10}^2(1-2x+x^2)=x\\ Solvingforaboveequation\\ 3.8\times {10}^2-2(3.8\times {10}^{2}x)+(3.8\times {10}^2)x^2=x\\ 3.8\times {10}^2-7.6\times {10}^{2}x+(3.8\times {10}^2)x^2=x\\ Collectingtermsweget,\\ (3.8\times {10}^2)x^2-7.6\times {10}^{2}x+3.8\times {10}^2=0\\ \text{Solving}\text{the}\text{quadratic}\text{equation}\text{gives}\text{the}\text{above}\text{equation}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ a=3.8\times {10}^2;b=-7.6\times {10}^{2}andc=3.8\times {10}^2\\ \text{The obtained values are substituted }\text{above equation}\\ x=\frac{-(-7.6\times {10}^2)\pm \sqrt{{(-7.6\times {10}^2)}^2-4(3.8\times {10}^2)(3.8\times {10}^2)}}{2(3.8\times {10}^2)}\\ x=\frac{760\pm \sqrt{{(760)}^2-4(1520)(3.8\times {10}^2)}}{760}\\ x=1.05(or)x=0.95\end{array}$

The second of these values 0.95 makes sense because concentration cannot be a negative number ( $\text{(1-x)=1}\text{.05;}\text{(1-1}\text{.05)=-0}\text{.5}$). Using the calculated values of (x), we can determine the equilibrium concentration of each species as fallowed below.

$\begin{array}{l}\text{[OD17X]}=0.95\text{}M,\\ \text{[OD1A][OF2A]}=(1-0.95)=0.05M\end{array}$

The each reactant and product molar concentrations are derived given the biochemical equilibrium reactions.

Conclusion

The each of reactant and product equilibrium concentration values are derived given the biochemical equilibrium reactions with respective statements.