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6.E: Chemical Equilibrium (Exercises)

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    41332
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    Q6.1

    Imagine a catalytic reaction that changes a reagent between two states, i.e.:

    \[yY + cC \rightleftharpoons cC + zZ\]

    The equilibrium constants \(K_c\), \(K_{\chi}\), and \(K_p\) with respect to concentration, mole fraction, and pressure are given by

    \[K_c = \dfrac{[Z]^z}{[Y]^y}\]

    \[K_{\chi} = \dfrac{\chi_Z^z}{\chi_Y^y}\]

    \[K_p = \dfrac{P_Z^z}{P_Y^y}\]

    Given this information, express \(K_p\) in terms of \(K_c\) and \(K_{\chi}\).

    Q6.2a

    \[CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(s)}\]

    1. The partial pressure of CO2 is 0.1 atm, calculate \(K_p\)
    2. If the reaction occurred in a 1L flask, what is the amount of CaO formed when 0.12 moles CaCO3 was set to react?
    3. What is the minimum about of CaO needed to cause the formation of CaCO3?

    Q6.2b

    The decomposition of sodium chlorate is:

    \[2NaClO_{3(s)} \rightleftharpoons 2NaCl_{(s)} + 3O_{2(g)}\]

    Suppose 0.760 mole of sodium chlorate was to be decomposed in a 3.75 L container. At the temperature 336°C, the percent decomposition of NaClO3 is 3.5%. Find the

    1. pressure of \(O_2\) gas in the container in atm and
    2. equilibrium constant \(K_p\) of the reaction. Assume \(O_2\) gas is an ideal gas.

    Hint: Use partial pressure and look at chemical equilibrium of gaseous systems

    S6.2b

    a) number of moles of NaClO3 decomposed is:

    \[ \dfrac{ \# moles \, decomposed}{.760 moles} = 0.035 \Rightarrow \# mole\,decomposed = 0.0266 mole \,NaClO_{3} \]

    number of moles of O2 formed is:

    \[ 0.0266 mol \,NaClO_{3} \times \dfrac{3 mol O_{3}}{2 mol \,NaClO_{3}} = 0.0399 mol \,O_{2} \]

    \[ P_{O_{2}} V = nRT \Rightarrow P_{O_{2}} = \dfrac{nRT}{V} \]

    \[ P_{O_{2}} = \dfrac{(0.0399 mol)(0.08206 \dfrac{L \cdot K}{K \cdot mol})(336+273)K}{3.75 L} = {\color{red} 0.532 atm} \]

    b) \[K_{p} = \dfrac{P_{O_{2}}}{P^{\circ}} = \dfrac{0.532 atm \times \dfrac{1.013 bar}{1atm}}{1bar} = {\color{red} 0.539 } \]

    Q6.2c

    Using the decomposition reaction of hydrogen peroxide (H2O2), assume that at 1048 °C and the pressure of the oxygen gas (O2) is 1.5 bar.

    \[\ce{2H2O2 (aq) \rightleftharpoons 2H_2O (l) + O2 (g) }\]

    1. Determine the \(K_P\) for the reaction above.
    2. If 0.92 mole of \(H_2O_2\) is placed in a 3.0-L beaker at 1048°C, determine the fraction of \(H_2O_2\) that will decompose.
    3. Determine the same as (b) if 1.3 moles of \(H_2O_2\) were placed inside the beaker instead.
    4. What is the minimum amount of \(H_2O_2\) (in moles) necessary in order to reach equilibrium?

    S6.2c

    a. CodeCogsEqn-14.gif

    b. In order to calculate the number of moles of H2O2 that are decomposed, you first need to determine the moles of O2 formed by the reaction. Treat O2 as an ideal gas.

    CodeCogsEqn-17.gif

    CodeCogsEqn-18.gif CodeCogsEqn-20.gif

    CodeCogsEqn-21.gif

    CodeCogsEqn-22.gif

    c. If 1.3 moles of H2O2 were used instead of 0.92 mole, the pressure of the O2 would not be affected and would remain at 1.5 bar. The number of moles of H2O2 decomposed would still be 0.0802 mole, therefore, the fraction of H2O2 decomposed would be as follows:

    CodeCogsEqn-23.gif

    d. The pressure of O2 must be greater than or equal to 1.5 bar in order for equilibrium to take place, therefore, the number of moles of H2O2 cannot be less than 0.0802 mol.

    Q6.3

    Find the value of Kp for cellular respiration if \(P_{O_{2}}=350torr\), with a 3:2 ratio to the pressure of CO2

    S6.3

    \[C_6H_{12}O_6(s)+O_2(g)\rightleftharpoons CO_2(g)+H_2O(l)\]

    \[P_{CO_{2}}=(350torr)\dfrac{2}{3}=233.3torr\]

    \[K_p=\dfrac{\dfrac{233.3torr}{750torr}}{\dfrac{350torr}{750torr}}=0.666\]

    Q6.5

    For the reaction of

    \[\ce{N2O4 → 2 NO2}\]

    with Kp = 0.167 at 300 K. If 1.0 g of N2O4 is placed into a 250.0 mL container:

    1. What would be its pressure if none dissociates?
    2. What is its partial pressure at equilibrium (with dissociation)?
    3. What is the total pressure in the container?

    Q6.9a

    Calculate \(\Delta_{vap}H\) for the evaporation of methanol when the temperature is raised from 20 °C to 100 °C and if the \(K_2/K_1\) ratio is 22.14?

    \[CH_3OH_{(l)} \rightarrow CH_3OH_{(g)}\]

    Q6.9b

    Using the decomposition reaction of coppr (II) oxide below:

    \[4CuO_{(s)} \rightleftharpoons 2Cu_2O_{(s)} + O_{2(g)}\]

    determine the standard enthalpy of the reaction. (Note: The equilibrium vapor pressures of O2 are 15.4 mmHg at 600°C and 927 mmHg at 850°C).

    S6.9b

    In this case, we must use the van't Hoff equation:

    \[ \ln \dfrac{K_2}{K_1} = \dfrac{\Delta_ H^o}{R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)\]

    in order to solve this particular problem. We can conclude that CodeCogsEqn-56.gif is proportional to CodeCogsEqn-57.gif due to the defined relationship of:

    CodeCogsEqn-58.gif .

    Therefore,

    CodeCogsEqn-59.gif

    CodeCogsEqn-60.gif

    CodeCogsEqn-61.gif

    Q6.10

    A container of water at 20ºC was placed in a freezer that was at a temperature of -5.0ºC. The vapor pressure of water in the container went from 0.60 bar to 0.38 bar. Calculate the enthalpy of fusion for the reaction that took place.

    S6.10

    • \(k_2=0.38 \;bar\)
    • \(k_1=0.60\; bar\)
    • \(T_2= 268\;K\)
    • \(T_1=293\;K\)

    \[\ln\dfrac{k_2}{k_1}= \dfrac{\Delta_r H^\circ }{R}\dfrac{T_2-T_1}{T_2T_1}\]

    \[\rightarrow \Delta_r H^\circ = \ln\dfrac{k_2}{k_1}(R)\dfrac{T_2T_1}{T_2-T_1}\]

    \[ \ln\dfrac{0.38 bar}{0.60 bar} 8.314 \dfrac{J}{mol\cdot K} \dfrac{(293\; K)(268\; K)}{268\; K-293\; K}(\dfrac{1\;kJ}{1000\; J})= 11.9 \dfrac{kJ}{mol}\]

    Q6.11

    The chemical responsible for the brown air throughout the Los Angeles area is NO2(g). To learn more about NO2(g), you decide to study this pollutant spectroscopically (by light absorption). You fill a gas cell with N2O, equilibrate the temperature to 298.0 K and then open the stopcock on the cell to equilibrate the pressure to the barometric pressure (723.4 mm Hg) that day. You then reequilibrate the cell at 323.1 K, 348.0 K, and 372.9 K. The following data is obtained:

    T/K 298.0 323.1 348.0 372.9
    [NO2]/M 0.01262 0.02140 0.02756 0.02920

    Find KP, Δ rxn G, Δ rxn H, and Δ rxn S at each temperature for the reaction

    \[\ce{N2O4(g) ⇌ 2 NO2(g)}.\]

    Q6.14

    What is the equilibrium constant and standard Gibbs energy change for:

    \[CO_{(g)} \rightleftharpoons CO_{(g)} + ½O_{2(g)}\]

    Assume that when \(CO_{(g)}\) dissociates into \(CO_{(g)}\) and \(O_{2(g)}\) at 300 K and 1.5 atm, the overall dissociation is one two fifths complete.

    Q6.15

    The standard Gibbs energies of formation of proban-1-ol and proban-2-ol are -171.3 kJ mol-1 and -180.3 kJ mol-1, respectively. Find the ratio of equilibrium vapor pressures of each isomer at 300 K.

    S6.15

    The ratio of equilibrium pressure is : 171.3/180.3 = 0.95

    Q6.16

    At what temperature does a particular reaction favor the formation of products at equilibrium if ΔrHº = 215.7 kJ mol-1 and ΔrSº = 348.8 J mol-1 K-1?

    S6.16

    The reaction favors the formation of products at equilibrium when

    \[\Delta_{r} G^{\circ} = \Delta_{r} H^{\circ}-T\Delta_{r} S^{\circ}<0\]

    \[\Delta_{r} H^{\circ}-T\Delta_{r} S^{\circ} = 215.7 \ast 10^{3} J\ mol^{-1} - T(348.8 J\ mol^{-1}\ K^{-1})<0\]

    \[T>\dfrac{215.7 \ast 10^{3} J\ mol^{-1}}{348.8 J\ mol^{-1}\ K^{-1}}\]

    \[T > 616 K\]

    Q6.17

    Using a table of thermodynamic data, find Ksp at 298.15K for:

    \[CO_{(g)} \rightarrow CO_{(g)} + ½O_{2(g)}\]

    (Hint, this problem can be solved using the van’t Hoff equation in its integrated form)

    Q6.18

    Consider the the Haber process:

    \[N_{2(g)} + 3H_{2 (g)} \rightleftharpoons 2NH_{3 (g)}\]

    How does the equilibrium shift under the following changes

    1. the volume of nitrogen gas is increased
    2. the temperature is decreased
    3. the pressure of the system is decreased

    S6.18

    1. The reaction shifts to the right : more products formed.
    2. The reaction shifts to the right : favor exothermic reaction
    3. The reaction shifts to the left : more pressure applied.

    Q6.19

    Given the general reaction A(g) ⇔2B(g), calculate the degree of dissociation of A at 25ºC and 7.00 bar if ΔrGº = 6.76 kJ mol-1. According to Le Chatelier's Principal, in what direction should this reaction proceed?

    S6.19

    \[\Delta_{r} G^{\circ} = -RTlnK_{P}\]

    \[K_{P} = e^{\dfrac{\Delta_{r} G^{\circ}}{-RT}} = e^{\dfrac{6.76\ast 10^{3}J\ mol^{-1}}{-(8.314J\ mol^{-1}\ K^{-1})298K}}\]

    \[K_{P} = 0.0653\]

    \[K_{P} = \dfrac{4\alpha ^{2}}{1-\alpha ^{2}}P\]

    \[\dfrac{4\alpha ^{2}}{1-\alpha ^{2}} = \dfrac{0.0653}{7}\]

    \[\alpha = .048\]

    Degree of dissociation α = 0.048, the reaction proceeds 4.8%. Le Chatelier's Principal says that reactions will move towards the side with fewer moles of gas at high pressures.

    Q6.19

    How does Le Chatelier's Principle relate to the following equation?

    CodeCogsEqn-66.gif

    S6.19

    Le Chatelier's Principle basically states that a system will adjust itself in efforts to re-establish equilibrium when an outside stress is placed on it. Using the equation provided, we can conclude that raising the temperature causes the equilibrium to shift from left to right in an endothermic reaction. This suggests that it is favoring the formation of products. We can also conclude that the opposite is true. This conclusion reinforces Le Chatelier's Principle because the temperature acts as the external stress placed on the system in this case.

    Q6.20

    H2 and H molecules are at equilibrium pressures of .35 bar and .30 bar, respectively. If the size of the container they are in is reduced by a factor of two, what will be the new partial pressures?

    S6.20

    \[H_2(g)\rightleftharpoons 2H(g)\]

    \[K_p=\dfrac{P_H^2}{P_{H_{2}}}=\dfrac{.30^2}{.35}=.675\]

    At new volume: \[P_H=.60\]

    \[P_{H_{2}}=.70\]

    Pressures will increase with decreased volume. Less molecules of gas will be favored.

    \(P_{H}\) \(P_{H_2}\)
    0.60 bar - 2x 0.70 bar + x

    \[.675=\dfrac{.60-(2x^2)}{.70+x}\rightarrow 4x^2 +.675x+.113\rightarrow x=.272\]

    \[P_{H_{2}}=.70bar+.272bar=.972bar\]

    \[P_H=.60bar-(2).272bar=.056bar\]

    Q6.21

    CodeCogsEqn (23).gif

    With Le Chatlier's principle in mind, does raising the temperature favor the forward reaction or the reverse?

    S6.21

    This reaction favors the reverse reaction.

    Q6.23a

    When a gas was heated at atmospheric pressure and 25°C, its color deepened. Heating above 150°C caused the color to fade, and at 550°C the color was barely detectable. At 550°C, however, the color was partially restored by increasing the pressure of the system. Which of the following scenarios best fits the above description? Justify your answer.

    1. A mixture of Gas A and Gas B
    2. Pure Gas B
    3. A mixture of Gas X and Gas Y

    (Hint: Gas B is bluish, and Gas Y is yellow. The other gases are colorless. Gas A and B are in their natural state, Gas AB has a ΔfH° = -43.2 kJ / mol. The reaction of Gas X into Gas Y is endothermic.)

    Q6.23b

    Hydrogen gas and iodine react at equilibrium in a glass canister to form Hydrogen Iodide, a strong acid:

    \[H_{2(g)}+I_{2(g)}\rightleftharpoons HI_{(g)}\]

    Iodine gas is a deep purple color. Both Hydrogen Iodide and hydrogen gas are colorless. Assume that iodine sublimates readily at 37o C, and that the reaction is endothermic in the direction written. What color is the gas mixture in the canister in the following scenarios? Provide an explanation for each.

    1. The canister is heated to 40o C
    2. The canister is heated to 500o C
    3. Negative pressure is applied to the cannister at 500o C

    Hint: At room temperature, the gas is colorless. Solid iodine must be in gas phase to react.

    S6.23b

    a.) The color of the gas will be a deep purple. The iodine will have sublimated, but the temperature is not high enough to drive the reaction forward.

    b.) The gas will be colorless, or almost completely so. This is because an increase in the temperature in an endothermic reaction drives the equilibrium constant higher. This can be justified using Le Chatelier's principle, which states that added stress to an equilibrium will be offset by the system.

    \[H_{2(g)}+I_{2(g)}+heat\rightleftharpoons HI_{(g)}\]

    Thus, when heat is added, the system will compensate by driving the reactants of the reaction forward into products. Another way to assess this is by the altered form of the van't Hoff equation:

    \[lnK=-\dfrac{\Delta _{r}H^{\circ}}{RT}+\dfrac{\Delta _{r}S^{\circ}}{R}\]

    Assuming that neither change in entropy nor enthalpy changes due to change in temperature, and that in an endothermic reaction the enthalpy is positive, as the temperature increases, so does K.

    c.) The gas will become slightly purple. This is because a decrease in pressure will alter K and drive the reaction in the direction which produces more moles of gas. By this, Le Chatelier's principle once again holds.

    Q6.24

    \[Mg + Pb^{2+} \rightarrow Mg^{2+}+Pb\]

    Write the two separate reactions happening here. Label which one is happening at the cathode end and which is happening at the anode end.

    S6.24

    Cathode:

    \[Pb^{2+}+2e^- \rightarrow Pb\]

    Anode:

    \[Mg \rightarrow Mg^{2+} + 2e^-\]

    Q6.24

    The reaction of N­2 (g) and H2 (g) gas produces NH3 (g). This reaction is exothermic explain what happens when you increase the temperature of the reaction. What happens when you increase the pressure?

    S6.24

    Since this reaction is exothermic heat is produced. Using Le Chatelier's principle we see that an increase in the temperature of the reaction will drive the reaction backward since heat is already in the product side. If we increase pressure we decrease the volume of the reaction therefore by Le Chatelier's principle an increase in the pressure should drive the reaction forward.

    Q6.25

    At places such as high mountain, the air pressure is lower than 1 atm, resulting lower partial pressure of Oxygen. What would you expect for the concentration of hemoglobin for the people living at such places?

    Q6.26

    Show the steps to get from

    \[Y = \dfrac{[PL]}{[L] + [P]}\]

    to

    \[\dfrac{1}{Y} = 1 + \dfrac{K_d}{L}\]

    Q6.27

    After an experiment of protein-binding you find the data respectively

    Total \(\dfrac{Mg^{2+}}{\mu M}\): 60 120 180 240 480

    \(\dfrac{Mg^{2+} bound to protein}{\mu M}\): 33.8 120 180 240 480

    Determine the dissociation constant of \(Ca^{2+}\) graphically. The protein concentration was kept at 96 \(\mu M\) for each run.

    Solution:

    Graph.PNG

    \[54.2\mu M\,pH=\,pKa\]

    \[54.2\mu M(\dfrac{1\times10^{6}M}{1\mu M})=5.42\times10^{-5}\]

    \[Ka=10^{-5.42E-5}\]

    \[Ka=.9998\]

    Q6.28

    The dissociation constant for the following reaction is 3.2x10-4. Dissolve 0.02M of C9H8O4 in water, calculate the molarity of reactants at equilibrium.

    C9H8O4 + H2O <---> H3O+ + C9H7O4-

    Q6.29

    The reaction

    \[ \begin{align} &Glyceraldehyde \ 3-phosphate + NAD^{+} + HPO_{4}^{2-} \rightarrow 1,3-Biphosphoglycerate + NADH + H^{+} \\ & \triangle_rG^{\circ'}=6.3 \ kJ \ mol^{-1} \end{align}\]

    is catalyzed by GAPDH (Glyceraldehyde 3-phosphate Dehydrogenase). At 298 K, predict whether or not the reaction will be spontaneous with the following information:

    [G3P] = 1.5x10-5 M; [BPG] = 3.0x10-3 M

    [NAD+] = 1.2x10-5 M; [NADH]=1.0x10-4 M

    [HPO42-]= 1.2x10-5 M; pH = 7.5

    S6.29

    Use the following equation.

    2.png

    Calculate [H+]

    \[ [H^{+}] = 10^{-7.5} = 3.162 X 10^{-8} M \\ since \ pH=-log([H^{+}])\]

    Plug values into equation and solve. Note: Since there is no coefficient in front of H+ in the reaction, x=1.

    \[\triangle_rG = 6.3 \ kJ mol^{-1} + (\dfrac{8.3145}{1000}kJ/mol^{-1} K^{-1}) (298K) \ln\dfrac{(3 \cdot 10^{-3} \ M)(3.162 \cdot 10^{-8} \ M/(1 \cdot 10^{-7})]^{1} \left(1.0 \cdot 10^{-4} \ M \right )}{ \left(1.5 \cdot 10^{-5} \ M \right ) \left(1.2 \cdot 10^{-5} \ M \right ) \left(1.2 \cdot 10^{-5} \ M \right )}\]

    \[ \underline {\triangle_rG = 49.90 \ kJ \ mol^{-1}}\]

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    Q6.29

    The reaction: \[Glucose+ATP\rightleftharpoons Glucose 6-phosphate + ADP\] At 298K, the equilibrium constant for the reaction is \(3.7 \times 10^{-3}\).Will the reaction occur spontaneously if the reaction is at the following concentrations:

    [Glucose]=\(3.2 \times 10^{-4}M\), [ATP]=\(2.5 \times 10^{-3}M\), [G-6-P]=\(1.2 \times 10^{-5}M\), [ADP]=\(1.0 \times 10^{-5}M\).

    S6.29

    \(\begin{align*}\Delta G^\circ &=-RTln(K_{eq})=-(8.314J/K*mol)(298K)ln(3.7 \times 10^{-3})\\&=13872.97J/mol=13.87297kJ/mol\end{align*}\)

    \(\begin{align*}
    \Delta G &=\Delta G^\circ +RTln(\dfrac{[Product]}{[Reactant]})\\&=13.87297kJ/mol + (8.314 \times 10^{-3}kJ/K*mol)(298K)ln(\dfrac{[1.2 \times 10^{-5}][1.0 \times 10^{-5}]}{[3.2 \times 10^{-4}][2.5 \times 10^{-3}]})\\&=-7.942kJ/mol
    \end{align*}\)

    \(\Delta G\) is negative, so the reaction is spontaneous at the given concentrations.

    Q6.30

    The established standard Gibbs energy for hydrolysis of ATP to ADP at 310K is \(-30.5kj\,mol^{-1}\). At \(-4.6^oC\), determine the \(\Delta rG^{o'}\) in the process of the muscle of a hippo. (Hint: \(\Delta rH^{o'}=-20.1kjmol^{-1}\)

    S6.30

    Step 1:

    \[\Delta rG^{o'}=\Delta rH^{o'}-T\Delta rS^{o'}\]

    Step 2:

    \[\Delta rS^{o'}=\dfrac{\Delta rH^{o'}-\Delta rG^{o'}}{T}$$
    $$=\dfrac{(-20.1kj\,mol^{-1})-(-3.0kj\,mol^{-1})}{310K}=3.355\times10^{-2}kj\,K^{-1}\,mol^{-1}\]

    Step 3:

    \[\Delta rG^{o'}=\Delta rH^{o'}-T\Delta rS^{o'}\]

    \[=(-20.1kj\,mol^{-1})-(298.15K)(3.355\times10^{-2}kj\,K^{-1}\,mol^{-1})\]

    $$=-29.1kj\,mol^{-1}$$

    Q6.31a

    Which step of glycolysis would not occur spontaneously at standard-state conditions and why?

    Q6.31b

    Consider a hydrolysis of PEP, a phosphate compound.

    PEP + H2O → pyruvate + Pi ΔrG°' = -61.9 kJ/mol

    At the temperature of 288K, the following reaction took place and has a reaction Gibbs free energy of -49.5 kJ/mole. Find the concentration of PEP for the reaction if the other concentrations are: [Pi] = 3.54 x 10-1 M, [Pyruvate] = 1.85 x 10-2 M.

    Hint: Use equation ΔrG° = -RTlnKp

    S6.31b

    \[\Delta _{r} G = \Delta _{r} G^{\circ '} + RTln\dfrac{[pyruvate][P_i]}{[PEP]}\]

    \[-49500 \dfrac{J}{mol} = -61900 \dfrac{J}{mol} + (8.314 \dfrac{J}{K \cdot mol})(288K)ln\dfrac{(1.85 \times 10^{-2}M)(3.54 \times 10^{-1}M)}{[PEP]}\]

    \[ {\color{red} [PEP] = 3.69 \times 10^{-5} M} \]

    Q6.32

    For the following reaction: \[Fructose 1,6-bisphosphate\rightleftharpoons dihydroxyacetone phosphate + glyceraldehyde 3-phosphate\] The \(\Delta G°'=5.7kcal/mol\). Calculate the equilibrium constant and determine if the reaction is spontaneous or not at 310K.

    S6.32

    \(\Delta G°'=5.7kcal/mol=23.8488kJ/mol \)

    \(\Delta G°'=-RTln(k_{eq}')\\k_{eq}'=e^{\dfrac{\Delta G°'}{-RT}}=e^{\dfrac{23848.8J/mol}{(-8.314J/K*mol)(310K)}}=9.58 \times 10^{-5}\)

    Small \(k_{eq}'\) suggest that this reaction is not a spontaneous process under the given conditions.

    Q6.32a

    Consider the following reaction:

    Glucose + Fructose → Sucrose + H2O

    Find:

    1. the value of the standard free energy at 300 K.
    2. the ratio between ∆rGo' at 300 K and ∆rGo' at 333 K.

    S6.32a

    a) The standard free energy = (-1544.3)- (-908.9 + -875.9) = 240.2 kJ/mol

    b)delta Go at 300 K = delta G1 + RTln Q = -34.6 kJ/mol

    delta GO at 333 K = delta G2 + RTln Q = -26.8 kJ/mol

    Q6.32b

    Consider the formation of the dipeptide glycylclycine. Using the following information, calculate the ΔrG.

    \[ \begin{align} & 2 \ Glycine \rightarrow Glycylglycine + H_2O \\ & \triangle _{r}G^{\circ ' }=29.5 \ kJ \ mol^{-1} \\ & \left [ Glycine \right ]= 1.4M\\ & \left [ Glycylglycine \right ]= 0.7M \\ & \left [ H_2O \right ]= 1.0M \end{align}\]

    S6.32b

    Use the following equation and plug in values.

    \[ \begin{align} & \triangle_rG^{\circ} = \triangle_rG^{\circ '} + RT\ln(K) \\ & \triangle_rG^{\circ} = \triangle_rG^{\circ '} + RT\ln\dfrac{[Glycylglycine]\cdot [H_2O]}{[Glycine]^{2}} \\ & \triangle_rG^{\circ} = 29.5 \ kJ \ mol^{-1} + (\dfrac{8.3145}{1000} \ kJ \ mol^{-1} \ K^{-1})(298K)\ln\dfrac{0.7M\cdot 1.0M}{1.4^{2} \ M} \\ & \underline{\triangle_rG^{\circ} = 26.7 \ kJ \ mol^{-1}} \end{align}\]

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    Q6.32c

    Consider the following reaction:

    \[2NH_{3(g)}+H_{2}O_{(l)}\rightleftharpoons 2NH^{+}_{4(aq)}\]

    \[\Delta _{f}\bar{G}^{\circ}(NH_{3(g)})=-16.6 \dfrac{KJ}{mol}\]

    \[\Delta _{f}\bar{G}^{\circ}(H_{2}O_{(l)})=-237.2 \dfrac{KJ}{mol}\]

    \[\Delta _{f}\bar{G}^{\circ}(NH^{+}_{4(aq)})=-79.3 \dfrac{KJ}{mol}\]

    What is the equilibrium constant of this process? Is this process typically spontaneous process under standard conditions?

    S6.32c

    To calculate change in Gibbs free energy for the equation, we use the standard molar Gibbs energies of formation for the reactants and products:

    \[\Delta _{r}G = 2\Delta _{f}\bar{G}^{\circ}(NH^{+}_{4(aq)})-\Delta _{f}\bar{G}^{\circ}(H_{2}O_{(l)})-2\Delta _{f}\bar{G}^{\circ}(NH_{3(g)})\]

    \[=2\left(-79.3\dfrac{kJ}{mol}\right)-\left(-267.2\dfrac{kJ}{mol}\right)-2\left(-16.6\dfrac{kJ}{mol}\right)\]

    \[=141.8 \dfrac{kJ}{mol}\]

    To calculate the equilibrium constant:

    \[\ln K=-\dfrac{\Delta _{r}G }{RT}=-\dfrac{141800\dfrac{J}{mol}}{(8.3145\dfrac{J}{K\cdot mol})(298K)}=-57.23\]

    \[K=e^{-57.23}=1.40 \times 10^{-25}\]

    Because the K value is so low for this process, the reaction does not occur spontaneously at 25o C

    Q6.33a

    Calculate the ΔrGº and equilibrium constant for the following reaction at 298K:

    3-Phosphoglycerate → Phosphoenolpyruvate +H2O

    Given that:

    2-Phosphoglycerate → 3-Phosphoglycerate ΔrGº = -4.2 kJ mol-1

    2-Phosphoglycerate → Phosphoenolpyruvate +H2O ΔrGº = -16.4 kJ mol-1

    S6.33a

    ΔrGº = 4.2 kJ mol-1 - 16.4 kJ mol-1 = 12.2 kJ mol-1

    \[\Delta _{r}G^{\circ} = -RTlnK\]

    \[lnK = \dfrac{12.2 \ast 10^{3} J\ mol^{-1}}{-(8.314J\ mol^{-1}\ K^{-1})298K}\]

    \[K = e^{.00492}\]

    \[K = 1.00\]

    Q6.33b

    Two acid dissociation reactions for carbonic and acetic acid and their corresponding pKa values are displayed below.

    CH3CO2H CH3CO2- + H+ ; pKa = 4.75

    2CO3 HCO3- + H+; pKa = 6.37

    Assuming the reactions take place at 298K, what is the equilibrium constant and change in gibbs’ free energy associated with the following reaction?

    CH3CO2- + H­2CO3 HCO3- + CH3CO2H

    Q6.34

    Suppose the isomerization of DHAP to GAP in glycolysis has an enthalpy of -1.20 kJ/mol. At 25°C the Gibbs free energy of the reaction is 1.98 kJ/mol. Determine the equilibrium constant K of the isomerization at 25°C and at 35°C

    Hint: Use van't Hoff equation

    S6.34

    For K at 25°C

    \[ \Delta _{r} G^{\circ} = -RTln K_{25^{\circ}} \]

    \[1980 \dfrac{J}{mol} = -(8.314 \dfrac{J}{K \cdot})(25+273)K (lnK_{25^{\circ}}) \Rightarrow {\color{red}K_{25^{\circ}} = 0.450}\]

    For K at 35°C

    \[ ln\dfrac{K_{2}}{K_{1}} = \dfrac{ \Delta _{r} H^{\circ}}{R}(\dfrac{1}{T_1} - \dfrac{1}{T_2}) \]

    \[ln\dfrac{K_{35^{\circ}}}{0.45} = \dfrac{-1200 J/mol}{8.314 \dfrac{J}{K \cdot mol}}(\dfrac{1}{(25+273)K} - \dfrac{1}{(35+273)K})\]

    \[{\color{red} K_{35^{\circ}} = 0.152 }\]

    Q6.35a

    Steady state and equilibrium state have an important role in understanding enzyme kinetics. What are some significant differences between the two?

    S6.35a

    In a steady state, there is no net change over time in concentrations of reactants and products of a reaction since they are being produced and consumed at constant rates. In this sense a steady state is a dynamic equilibrium. A steady state can be going in either the forward reaction or the backward reaction.

    A chemical equilibrium on the other hand is when a reaction goes in the forward and backward reaction at the same rate so there is no net change in the system.

    This is important because cells maintain steady states so that they are able to use particular reactions continuously. If a cell were at chemical equilibrium it would be dead because it would be at the point where all reactions are not going anymore, among other reasons.

    Also, it is important to note that in a steady state, reactions are reversible compared to chemical equilibrium where the reaction rate is zero.

    Click here to see more information about steady states.

    Q6.35b

    Is the following an example of a steady state, equilibrium state, or neither:

    1. Glycolysis, ingestion, and respiration
    2. NaCl_{(s)}\rightleftharpoons Na^{+}_{(aq)}+Cl^{-}_{(aq)}
    3. Sodium potassium ATPase
    4. Citric Acid Cycle
    5. oxidation of gold

    S6.35b

    1. steady state
    2. equilibrium state
    3. steady state
    4. steady state
    5. neither

    Q6.36

    Diatomic hydrogen gas and diatomic iodine gas are in equilibrium with hydrogen Iodide gas in a closed environment of unknown volume at an unknown temperature.

    Find the pressure equilibrium constant if the partial pressures of the gases are as follows: PH2 = 812 mmhg PI2= 587 mmhg PHI = 980 mmhg

    Find the activity equilibrium constant if the activity constants of the gases are as follows γH2 = 1.45 γI2=0.844 γHI = 1.23

    Use your results to find the thermodynamic equilibrium constant.

    S6.36

    The reaction in question is H2 + I2 ⇌ 2HI

    The pressure equilibrium constant is given by PHI2/(PH2 PI2), the activity equilibrium constant is given by γHI2/(γH2 γI2), and the thermodynamic equilibrium constant is given by the product of those two.

    Kp = (980 mmhg)2/(812 mmhg * 587 mmhg) = 2.01

    Kγ = 1.232/(1.45 * 0.844) = 1.24

    Kthermo = 1.24 * 2.01 = 2.49

    Q6.38

    \[SO_2 + Cl_2 \rightarrow SO_2Cl_2 \]

    This reaction happens at 273K. You are given a Kp of 0.683. The pressure for SO2 is 0.58 bar for Cl2 is 0.93 bar, and SO2Cl2 is 0.776 bar. From this information, determine (delta)rG.

    CodeCogsEqn (3).gif

    =(-8.314Jmol-1K-1)(273K)ln(0.683)

    =865.4 J mol-1

    CodeCogsEqn (4).gif

    CodeCogsEqn (25).gif

    =1.69 kj mol-1

    Q6.40

    Assuming oxygen binding to hemoglobin can be represented by the following reaction:

    \[Hb_{(aq)} + O_{2(g)} \rightarrow HbO_{2(aq)}\]

    If the value of ΔrG° for the reaction is -11.2 kJ mol-1 at 37°C, calculate the value of ΔrG° for the reaction.

    Q6.42

    At T=300K, given the mole ratio between 2 isomers Cis-2-butene and Trans-2-butene in an equilibrium mixture is 1:4. Evaluate rG of the reversible reaction:

    Cis-2-butene <------> Trans-2-butene

    Q6.44

    When discussing the reaction in biological cells, why would you use concentrations instead of activities?

    S6.44

    Concentrations are generally smaller and it's easier to compare the concentrations between two parts of a cell or between different molecules in a reaction whereas activity describes behavior.

    Q6.46

    The following data shows the oxygen binding concentration in snails. The protein concentration is 15mM. Find \(n\) and \(K_d\) by using the Scatchard plot

    \([O_2]_{total}\) 10 mM 14 mM 18 mM 22 mM 26 mM 30 mM 34 mM 38 mM 42 mM 46 mM
    \([O_2]_{bound}\) 9 mM 12 mM 16 mM 19 mM 22 mM 25 mM 28 mM 32 mM 35 mM 39 mM

    S6.46

    \(y=mx+b\\
    \dfrac{Y}{[L]}=(-\dfrac{1}{K_d})Y+\dfrac{n}{K_d}\\
    Y=\dfrac{[O_2]_{bound}}{[P]}\\.[L]=[O_2]_{free}\)

    O2 binding.JPG

    [O2]tot

    10 14 18 22 26 30 34 38 42 46
    [O2]bound 9 12 16 19 22 25 28 32 35 39
    [O2]free 1 2 2 3 4 5 6 6 7 7
    Y 0.6 0.8 1.067 1.267 1.467 1.667 1.867 2.133 2.333 2.6
    Y/[L] 0.6 0.4 0.833 0.422 0.367 0.333 0.311 0.356 0.333 0.371

    \(-\dfrac{1}{K_d}=-0.102\\
    K_d=9.804\\
    \dfrac{n}{K_d}=0.564\\
    n=K_d*0.564=5.529\)


    6.E: Chemical Equilibrium (Exercises) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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