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15: Equilibria

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    15.1 Chemical Equilibrium

    Textbook: Section 15.1

    15.2 Equilibrium Constant and Reaction Quotient

    Textbook: Section 15.2

    Exercise \(\PageIndex{2.a}\)

    Use the following reaction to answer

    \(H_{2}(g)+Br_{2}(l)\rightleftharpoons 2HBr(g)\)

    Which one of the following is the equilibrium constant expression for at 25°C?

    1. \(k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]\left [ Br_{2} \right ]} \)
    2. \(k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]} \)
    3. \(k=\frac{\left [ HBr \right ]}{\left [ H_{2} \right ]}\)
    4. none of the above
    Answer

    b. \[k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]} \nonumber \]

    You do not include the Br2(l) because the concentration of pure incompressible states of matter (liquids and solids) are always a constant value (related to their density). Therefore, they are absorbed into the equilibrium constant, (a constant divided or multiplied by another constant is still an constant).

     

    Exercise \(\PageIndex{2.b}\)

    Use the following reaction to answer

    \(H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g)\)

    What is the equilibrium concentration of HBr if H2 and Br2 are both 0.800M, and K=8.0  at 70°C.

    Answer

    \[k=\frac{\left [ HBr \right ]^{2}}{\left [ H_{2} \right ]\left [ Br_{2} \right ]} \nonumber \]

    \[\left [ HBr \right ]=\sqrt{k\left [ H_{2} \right ]\left [ Br_{2} \right ]}=\sqrt{8.0*0.80^{2}}=2.26M \nonumber \]

     

    Exercise \(\PageIndex{2.c}\)

    Why is Br2 a liquid in question 15.2.a and a gas in 15.2.b ?

    Answer

    The boiling point of Br2 is 58.8°C. In question 15.2.a the temperature is below the boiling point, and in 15.2.b it is above.

     

    Exercise \(\PageIndex{2.d}\)

    Use the following reaction to answer

    \(H_{2}(g)+Br_{2}(l)\rightleftharpoons 2HBr(g)\)

    Calculate Kp at 25° for a system at equilibrium if: PH2 = 2.50x10-2 atm, and PHBr = 1.500atm?

    Answer

    \[K_{p}=\frac{\left ( P_{HBr} \right )^{2}}{P_{H_{2}}}=\frac{1.50^{2}}{2.50*10^{-2}}=90 \nonumber \]

     

    Exercise \(\PageIndex{2.e}\)

    Use the following reaction to answer

    \(H_{2}(g)+Br_{2}(l)\rightleftharpoons 2HBr(g)\)

    What is Kc for the system in question 15.3?

    Answer

    \[K_{c}=\frac{\left ( P_{HBr}/RT \right )^{2}}{P_{H_{2}}/RT}=\frac{\left ( P_{HBr} \right )^{2}}{P_{H_{2}}}*\left ( RT \right )^{-1} \nonumber   \]

    \[K_{c}=90*\left ( 0.08206*298.2 \right )^{-1} \nonumber \]

    \[K_{c}=3.68 \nonumber \]

     

    Exercise \(\PageIndex{2.f}\)

    \[2NO(g)\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} N_{2}O_{2}(g),\]

    \[N_{2}O_{2}(g)+O_{2}(g) \rightarrow 2NO_{2}(g)\]

    \[2NO(g)+O_{2}(g) \rightarrow 2NO_{2}(g)\]

    \[k_{1}=4.8*10^{-3}M^{-1}s^{-1},k_{-1}=3.6*10^{-6}s^{-1},k_{2}=6.4*10^{3}\nonumber\]

    a. What is the value of Kc for reaction 15.1 above?

    b.What is the value of Kc for reaction 15.3 above?

    c. Does reaction 15.1 favor reactants or products?

    Answer a

    \[K=\frac{k_{f}}{k_{r}}=\frac{k_{1}}{k_{-1}}=\frac{4.8*10^{-3}}{3.6*10^{-6}}=1.3333*10^{3}=1.3*10^{3}\nonumber\]

    Answer b

    \[K_{3}=K_{1}*K_{2}=\left ( 1.3333*10^{3} \right )\left ( 6.4*10^{3} \right )=8.5*10^{6}\nonumber\]

    Answer c

    \[K=\frac{k_{1}}{k_{-1}}=\frac{4.8*10^{-3}}{3.6*10^{-6}}=1.3*10^{3}\nonumber\]

    K>>1, the forward reaction is in favor, products

     

    15.3 Calculations of Equilibrium Constant

    Textbook: Section 15.3

    Exercise \(\PageIndex{3.a}\)

    \[H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber \]

    In what direction will the reaction at 440°C proceed if:

    [HBr]= 1.0x10-2M, [H2]=5.0x10-3M, [Br2]=1.5x10-2M and Keq=50 at 440°C.

    Answer

    \[Q=\frac{\left [ 1.0*10^{-2} \right ]^{2}}{\left [ 5.0*10^{-3} \right ]\left [ 1.5*10^{-2} \right ]}=1.3\nonumber\]

    Q < K, therefore, the reaction will proceed towards the product direction

     

    Exercise \(\PageIndex{3.b}\)

    \[H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber \]

    What is the equilibrium concentration of Bromine at 440°C if:

    [H2]= 0.60M, [HBr]=1.25M and [Br2]=0

    Answer
    R H2 (g)            + Br2 (g) 2HBr (g)
    I 0.60 M 0   1.25 M
    C x x   -2x
    E 0.60+x x   1.25-2x

    \[K=\frac{\left [ 1.25-2x \right ]^{2}}{\left [ 0.60+x\right ]\left [ x \right ]}=50 \nonumber \]

    \[\left ( 1.25-2x \right )^{2}=50\left ( 0.60+x^{2} \right ) \nonumber \]

    \[46x^{2}+35x-1.5625=0 \nonumber \]

    \[x=0.042M \nonumber \]

     

    Exercise \(\PageIndex{3.c}\)

    \[H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber \]

    What is [HBr] after equilibrium is reached for the problem described in question 15.a?

    Answer
    R H2 (g)            + Br2 (g) 2HBr (g)
    I 0.60 M 0   1.25 M
    C x x   -2x
    E 0.60+x x   1.25-2x

    \[K=\frac{\left [ 1.25-2x \right ]^{2}}{\left [ 0.60+x\right ]\left [ x \right ]}=50 \nonumber \]

    \[\left ( 1.25-2x \right )^{2}=50\left ( 0.60+x^{2} \right ) \nonumber \]

    \[46x^{2}+35x-1.5625=0 \nonumber \]

    \[x=0.042M \nonumber \]

    \[1.25-\left( 2*0.042\right) =1.166M \nonumber \]

     

     

    Exercise \(\PageIndex{3.d}\)

    \[H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber \]

    Determine Keq at a certain temperature if 0.80 mol HBr is produced after 0.50 mol H2 and Br2 react in a 2 L container.

    Answer
    R H2 (g)            + Br2 (g) 2HBr (g)
    I 0.25 M 0.25 M   0
    C -x -x   +2x
    E 0.25-x 0.25-x   2x

    \[0.8mol/2.0L=0.40M\nonumber \]

    \[0.40M=2x \nonumber \]

    \[x=0.2M \nonumber \]

    \[K=\frac{\left [ 0.4 \right ]^{2}}{\left [ 0.05 \right ]\left [ 0.05 \right ]}=64 \nonumber \]

     

    Exercise \(\PageIndex{3.e}\)

    \[H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber \]

    At a certain temperature, Keq=6.30x102. What is the equilibrium concentration of H2 if 4.50mol of each of the three species were placed into a 3.00L flask?

    Answer
    R H2 (g)            + Br2 (g) 2HBr (g)
    I 1.5 M 1.5 M   1.5 M
    C -x -x   +2x
    E 1.5-x 1.5-x   1.5+2x

    \[K=\frac{\left [ 1.5+2x \right ]^{2}}{\left [ 1.5-x \right ]\left [ 1.5-x \right ]}=6.3*10^{2} \nonumber \]

    \[x=1.33 M \nonumber \]

    \[\left [ H_{2} \right ]_{Equil}=1.5-1.33=0.17 M \nonumber \]

     

     

    Exercise \(\PageIndex{3.f}\)

    \[H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g) \nonumber \]

    What is [HBr] after equilibrium is reached for the problem described in question 15.d?

    Answer
    R H2 (g)            + Br2 (g) 2HBr (g)
    I 1.5 M 1.5 M   1.5 M
    C -x -x   +2x
    E 1.5-x 1.5-x   1.5+2x

    \[K=\frac{\left [ 1.5+2x \right ]^{2}}{\left [ 1.5-x \right ]\left [ 1.5-x \right ]}=6.3*10^{2} \nonumber \]
    \[x=1.33 M \nonumber \]
    \[1.5+\left( 2*1.33\right) = 4.16 M \nonumber \]

     

    Exercise \(\PageIndex{3.g}\)

    Determine Keq for the following reaction if 0.0500M N2O4 is placed in a container and it decomposes to an equilibrium value of 0.0155M.

    \(N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)\)

    Answer
    R N2O4 (g) 2NO2 (g)
    I 0.05 M   0
    C -x   +2x
    E 0.0155M   1.5+2x

    \[x=0.05-0.0155=0.0345M \nonumber \]

    \[K=\frac{\left [ 2x \right ]^{2}}{\left [ 0.0155 \right ]}=\frac{\left [ 2*0.0345 \right ]^{2}}{\left [ 0.0155 \right ]}=0.307\nonumber \]

     

    15.6 Le Chatelier’s Principle

    Textbook: Section 15.6

    \(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol\)

    Use this reaction to answer questions 6.a-f

    Exercise \(\PageIndex{6.a}\)

    \(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol\)

    If the temperature increases while the pressure is constant, the reaction will proceed towards which direction?

    Answer

    Reactants, because of \(\Delta H<0\)

     

    Exercise \(\PageIndex{6.b}\)

    \(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol\)

    For the same reaction, if the temperature is held constant and the pressure is increased, which direction will the reaction proceed?

    Answer

    Products, increasing the pressure favors the side with fewer gas molecules

     

    Exercise \(\PageIndex{6.c}\)

    \(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol\)

    If more oxygen is added, which direction will the reaction proceed?

    Answer

    Products, by adding more reactant this favors the formation of more product

     

    Exercise \(\PageIndex{6.d}\)

    \(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol\)

    If the container in which the reaction occurs is enlarged, which direction will the reaction proceed?

    Answer

    Reactants, by enlarging the container the partial pressure decreases, therefore it favors the side with more gas molecules

     

    Exercise \(\PageIndex{6.e}\)

    \(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol\)

    The addition of catalyst will make the reaction shift towards which direction?

    Answer

    There will be no effect, catalysts only change the activation energy of a reaction

     

    Exercise \(\PageIndex{6.f}\)

    \(2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3} (g)\,\,\,\,\,\Delta H^{0}=-99\,kJ/mol\)

    The addition of He gas will make the reaction shift towards which direction?

    Answer

    There will be no effect, the participation of an inert gas does not affect the reaction

     

    15.4 Equilibrium Constant

    Textbook: Section 15.4

    Exercise \(\PageIndex{4.a}\)

    \(C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)\)

    What is the concentration of C4H10 at equilibrium if the concentrations of C2H6 and C2H4 are both 0.014M? Kc = 0.07

    Answer

    \[K_{c}=\frac{\left [ 0.014 \right ]^{2}}{\left [ C_{4}H_{10} \right ]}=0.070 \nonumber\]

    \[\left [ C_{4}H_{10} \right ]=0.0028M \nonumber\]

     

    Exercise \(\PageIndex{4.b}\)

    \(C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)\)

    For the same reaction, if the initial concentration of C4H10 is 0.035M, and there is no C2H6 C2H4 present initially. What is the equilibrium concentration of C4H10?

    Answer
    R C4H10(g) C2H6(g) + C2H4(g)
    I 0.035 M   0 0
    C -x   +x +x
    E 0.035-x   x x

    \[K=\frac{\left [ x \right ]^{2}}{\left [ 0.035-x \right ]}=0.070 \nonumber \]

    \[x=0.026 M \nonumber \]

    \[0.035-0.026=0.009 M \nonumber\]

     

    Exercise \(\PageIndex{4.c}\)

    \(C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)\)

    Following Q 15.b, what is the equilibrium concentration of C2H6?

    Answer
    R C4H10(g) C2H6(g)     + C2H4(g)
    I 0.035 M   0 0
    C -x   +x +x
    E 0.035-x   x x
    \[K=\frac{\left [ x \right ]^{2}}{\left [ 0.035-x \right ]}=0.070 \nonumber\]\[x=0.026 M \nonumber\]
     

     

    Exercise \(\PageIndex{4.d}\)

    \(C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)\)

    If the initial concentration of C4H10 is 0.030M, and the ones of C2H6 and C2H4 are both 0.023M. At equilibrium, the concentration of C4H10 becomes 0.018M, what is the value of Kc of the reaction.

    Answer
    R C4H10(g) C2H6(g)     + C2H4(g)
    I 0.030 M   0.023 M 0.023 M
    C -x   +x +x
    E 0.018   0.023+x 0.023+x

    \[x=0.030-0.018=0.012 \nonumber\]

    \[K_{c}=\frac{\left [ 0.035 \right ]^{2}}{\left [ 0.018 \right ]}=0.070 \nonumber\]

    \[K_{c}=0.068 \nonumber\]

     

    Exercise \(\PageIndex{4.e}\)

    \(C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)\)

    For the same reaction, if the 2.0L container was evacuated, then pumped in with gases at the following pressure, C4H10 = 1.2atm, C2H6and C2H4 = 0.6 atm. What is the partial pressure (in atm) of C4H10 at equilibrium? Kp =0.64.

    Answer
    R C4H10(g) C2H6(g)     + C2H4(g)
    I 1.2 atm   0.6 atm 0.6 atm
    C -x   +x +x
    E 1.2-x   0.6+x 0.6+x

    \[K_{p}=\frac{\left [ 0.6+x \right ]^{2}}{\left [ 1.2-x \right ]}=0.64 \nonumber\]

    \[x=0.2 atm  \nonumber\]

    \[1.2-0.2=1.0atm \nonumber\]

     

    Equilibria Calculations Through "Completing the Power"

     

    Exercise \(\PageIndex{a}\)

    \(H_{2}(g)+Br_{2}(l)\rightleftharpoons 2HBr(g)\)

    For the reaction, what is the equilibrium concentration of Br2 at 4400C if initially [H2]= 0.60M [HBr]=1.25M [Br2]=0.6 and k=50?

    Answer
    R H2(g)            + Br2(g) 2HBr
    I 0.60 M 0.06 M   1.25 M
    C -x -x   +2x
    E 0.60-x 0.60-x   1.25+2x

    \[K=\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]\left [ 0.60-x \right ]}=50 \nonumber \]

    \[\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]^{2}}=50\nonumber \]

    \[\left (\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}\right )^{2}=50\nonumber \]

    \[\sqrt{\left (\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}\right )^{2}}=\sqrt{50}\nonumber \]

    \[\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}=7.07\nonumber \]

    \[x=0.33\,M\nonumber \]

    \[0.60-0.33=0.27\,M\nonumber \]

     

    Exercise \(\PageIndex{b}\)

    \(H_{2}(g)+Br_{2}(l)\rightleftharpoons 2HBr(g)\)

    What is the equilibrium concentration of HBr?

    Answer
    R H2(g)            + Br2(g) 2HBr
    I 0.60 M 0.06 M   1.25 M
    C -x -x   +2x
    E 0.60-x 0.60-x   1.25+2x

    \[K=\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]\left [ 0.60-x \right ]}=50 \nonumber \]

    \[\frac{\left [ 1.25+2x \right ]^{2}}{\left [ 0.60-x \right ]^{2}}=50\nonumber \]

    \[\left (\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}\right )^{2}=50\nonumber \]

    \[\sqrt{\left (\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}\right )^{2}}=\sqrt{50}\nonumber \]

    \[\frac{\left [ 1.25+2x \right ]}{\left [ 0.60-x \right ]}=7.07\nonumber \]

    \[x=0.33\,M\nonumber \]

    \[1.25+\left(2*0.33\right)=1.91\,M\nonumber \]

     

    Exercise \(\PageIndex{c}\)

    \(Cl_{2}(g)+2HBr(g)\rightleftharpoons Br_{2}(g)+2HCl(g)\)

    For the reaction, what is the equilibrium concentration of HCl if the initial [Cl2]=0.1M, [HBr]=0.20M, [Br2]=[HCl]=0 and k=15?

    Answer
    R Cl2(g)            + 2HBr(g) Br2(g) 2HCl(g)
    I 0.10M 0.20M   0 0
    C -x -2x   +x +2x
    E 0.10-x 0.20-2x   x 2x

    \[K=\frac{\left [ Br_{2} \right ]\left [ HCl \right ]^{2}}{\left [ HBr \right ]^{2}\left [ Cl \right ]}=\frac{\left [ x \right ]\left [ 2x \right ]^{2}}{\left [ 0.2-2x \right ]^{2}\left [ 0.1-x \right ]}=\frac{\left [ x \right ]\left(4\left [ x \right ]^{2}\right)}{\left(2\left [ 0.1-x \right ]^{2}\right)\left [ 0.1-x \right ]}\nonumber \]

    \[K=\frac{4\left [ x \right ]^{3}}{4\left [ 0.1-x \right ]^{3}}=\left (\frac{\left [ x \right ]}{\left [ 0.1-x \right ]}\right )^{3}\Rightarrow (K)^\frac{1}{3}=\frac{\left [ x \right ]}{\left [ 0.1-x \right ]}\nonumber \]

    \[(K)^\frac{1}{3}\left [ 0.1-x \right ]=\left [ x \right ]\Rightarrow -\left [ x \right ]\left ( 1+K^{1/3} \right )=-0.1K^{1/3}\nonumber \]

    \[\left [ x \right ]=\frac{0.1K^{1/3}}{K^{1/3}+1}=\frac{0.1*15^{1/3}}{15^{1/3}+1}=0.0712\nonumber \]

    \[2x=0.14\nonumber \]

     

    Exercise \(\PageIndex{d}\)

    \(Cl_{2}(g)+2HBr(g)\rightleftharpoons Br_{2}(g)+2HCl(g)\)

    What is the equilibrium concentration of HBr in Q15.29?

    Answer
    R Cl2(g)            + 2HBr(g) Br2(g) 2HCl(g)
    I 0.10M 0.20M   0 0
    C -x -2x   +x +2x
    E 0.10-x 0.20-2x   x 2x

    \[K=\frac{\left [ Br_{2} \right ]\left [ HCl \right ]^{2}}{\left [ HBr \right ]^{2}\left [ Cl \right ]}=\frac{\left [ x \right ]\left [ 2x \right ]^{2}}{\left [ 0.2-2x \right ]^{2}\left [ 0.1-x \right ]}=\frac{\left [ x \right ]\left(4\left [ x \right ]^{2}\right)}{\left(2\left [ 0.1-x \right ]^{2}\right)\left [ 0.1-x \right ]}\nonumber \]

    \[K=\frac{4\left [ x \right ]^{3}}{4\left [ 0.1-x \right ]^{3}}=\left (\frac{\left [ x \right ]}{\left [ 0.1-x \right ]}\right )^{3}\Rightarrow (K)^\frac{1}{3}=\frac{\left [ x \right ]}{\left [ 0.1-x \right ]}\nonumber \]

    \[(K)^\frac{1}{3}\left [ 0.1-x \right ]=\left [ x \right ]\Rightarrow -\left [ x \right ]\left ( 1+K^{1/3} \right )=-0.1K^{1/3}\nonumber \]

    \[\left [ x \right ]=\frac{0.1K^{1/3}}{K^{1/3}+1}=\frac{0.1*15^{1/3}}{15^{1/3}+1}=0.0712\nonumber \]

    \[0.20-2x=0.06M\nonumber \]

     

    Exercise \(\PageIndex{e}\)

    For the reaction, What is [O2] at equilibrium if 0.050mol of N2O2, O2, and NO2 are mixed in a 1.0L container? k=25

    \(N_{2}O_{2}(g)+O_{2}(g)\rightarrow 2NO_{2}(g)\)

    Answer
    R N2O2(g)            + O2(g) 2NO2(g)
    I 0.050 M 0.050 M   0.050 M
    C -x -x   +2x
    E 0.050-x 0.050-x   0.050+x

    \[K=\frac{\left [ 0.05+2x \right ]^{2}}{\left [ 0.05-x \right ]\left [ 0.05-x \right ]}=25\nonumber \]

    \[\frac{\left [ 0.05+2x \right ]^{2}}{\left [ 0.05-x \right ]^{2}}=25\nonumber \]

    \[\sqrt{\frac{\left [ 0.05+2x \right ]^{2}}{\left [ 0.05-x \right ]^{2}}}=\sqrt{25}\nonumber \]

    \[\frac{\left [ 0.05+2x \right ]}{\left [ 0.05-x \right ]}=25\nonumber \]

    \[x=0.03 M\nonumber \]

    \[0.05-0.03=0.02 M\nonumber \]

     

    Relating Kp and Kc

    Exercise \(\PageIndex{a}\)

    What is the relationship between Kp and Kc?

    1. \(K_{p}=K_{c}\left ( RT \right )^{\Delta n}\)     
    2. \(K_{c}=K_{p}\left ( RT \right )^{\Delta n}\) 
    3.  \(K_{p}=K_{c}\) 
    4. None of the above
    Answer

    a. \(K_{p}=K_{c}\left ( RT \right )^{\Delta n}\)

     

    Exercise \(\PageIndex{b}\)

    \(3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)\)

    What is the Δn for the equation above?

    Answer

    \[3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)\nonumber \]

    \[\Delta n=6-(3+2)\nonumber \]

    \[\Delta n=6-5=1\nonumber \]

     

    Exercise \(\PageIndex{c}\)

    \(3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)\)

    Solve for Kp, given Kc = 2.3*204 at 30°C for the equation above?

    Answer

    \[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber \]

    \[K_{p}=2.3*10^{4}\left (0.08206*303\right )^{\left(6-\left (3+2\right )\right)}\nonumber \]

    \[K_{p}=5.7*10^{5}\nonumber \]

     

    Exercise \(\PageIndex{d}\)

    \(3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)\)

    Solve for Kc, given Kp = 2.3*104 at 30°C for the equation above?

    Answer

    \[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber \]

    \[2.3*10^{4}=K_{c}\left (0.08206*303\right )^{\left (6-3\right )}\nonumber \]

    \[K_{c}=\frac{2.3*10^{4}}{\left (0.08206*303\right )^{\left (6-3\right )}}\nonumber \]

    \[K_{c}=1.5\nonumber \]

     

    Exercise \(\PageIndex{e}\)

    \(3A_{2}(g)+2B_{3}(s)\rightarrow 6A_{3}B_{2}(g)\)

    Solve for Kc, given Kp = 3.1*104 at 30°C for the equation above?

    Answer

    \[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber \]

    \[3.1*10^{4}=K_{c}\left (0.08206*298\right )^{\left (6-3\right )}\nonumber \]

    \[K_{c}=\frac{3.1*10^{4}}{\left (0.08206*298\right )^{\left (6-3\right )}} \nonumber \]

    \[K_{c}=2.1\nonumber \]

     

    Exercise \(\PageIndex{f}\)

    Solve for Kc, given Kp = 3.2*104 at 30°C for the following equation:

    \[3A_{2}(g)+4B_{3}(g)\rightarrow 2A_{3}B_{2}(g)\nonumber \]

    Answer

    \[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber \]

    \[3.2*10^{4}=K_{c}\left (0.08206*303\right )^{\left (2-4\right )}\nonumber \]

    \[K_{c}=\frac{3.2*10^{4}}{\left (0.08206*303\right )^{\left (2-4\right )}}\nonumber \]

    \[K_{c}=2.0*10^{7}\nonumber \]

     

    Exercise \(\PageIndex{g}\)

    What is the temperature, given Kc = 3.5*108, and Kp = 5.8*105 for the following equation?

    \[A_{2}(g)+3B_{3}(g)\rightarrow 2AB_{3}(g)\nonumber \]

    Answer

    \[K_{p}=K_{c}\left ( RT \right )^{\Delta n}\nonumber \]

    \[5.8*10^{5}=3.5*10^{8}\left (0.08206*T\right )\nonumber \]

    \[T=\frac{\left ( \frac{5.8*10^{5}}{3.5*10^{8}} \right )^\frac{1}{\left ( 2-4 \right )}}{0.08206}\nonumber \]

    \[T=299K\nonumber \]

     

    General Questions 

    Exercise \(\PageIndex{a}\)

    Which of the following is the relationship between the rate constants for the forward and reverse reactions and the equilibrium constant for a process?

    1. \(K= k_{f} + k_{r}\)
    2. \(K= k_{f}k_{r}\)
    3. \(K=k_{f}-k_{r}\)
    4. \(K= \frac{1}{\left ( k_{f}k_{r} \right )}\)
    5. \(K=\frac{k_{f}}{k_{r}}\)
    Answer

    e. \(K=\frac{k_{f}}{k_{r}}\)

     

    Exercise \(\PageIndex{b}\)

    A flask of an aqueous equilibrium mixture of \(CoCl_{4}^{2-}\), \(CoBr_{4}^{2-}\), \(Cl^{-}\), and \(Br^{-}\) is at 25°C. Which of the following actions will change the value of the equilibrium constant from that which currently describes the concentration relationships of the four species above?

    1. add more \(CoBr_{4}^{2-}\)  to the solution
    2. add more \(CoCl_{4}^{2-}\)  to the solution
    3. add more \(Br^{-}\)  to the solution
    4. add more \(Cl^{-}\)  to the solution
    5. put the flask into an 80°C water bath
    Answer

    e. put the flask into an 80°C water bath

     

    Exercise \(\PageIndex{c}\)

    What is the Kc for the following reaction?

    \(CO+3H_{2}\rightleftharpoons CH_{4}+H_{2}O\)

    Answer

    \[K_{c}=\frac{\left [ CH_{4} \right ]\left [ H_{2}O \right ]}{\left [ CO \right ]\left [ H_{2} \right ]^{3}}\nonumber \]

     

    Exercise \(\PageIndex{d}\)

    If the equilibrium constant for reaction (1) is 4.22*10-3, what is the value of the equilibrium constant for the reaction (2) in the following mechanism?

    3A + 2B ⇌ 2D + E    (1)

    2D + E ⇌ 3A +2B     (2)

    Answer

    \[k_{eq1}=\frac{\left [ D \right ]^{2}\left [ E \right ]}{\left [ A \right ]^{3}\left [ B \right ]^{2}}\nonumber \]
    \[k_{eq2}=\frac{\left [ A \right ]^{3}\left [ B \right ]^{2}}{\left [ D \right ]^{2}\left [ E \right ]}\nonumber \]
    \[k_{eq2}=\frac{\left [ A \right ]^{3}\left [ B \right ]^{2}}{\left [ D \right ]^{2}\left [ E \right ]}=\frac{1}{k_{eq1}}=\frac{1}{4.22*10^{-3}}=237\nonumber \]

     

    Exercise \(\PageIndex{e}\)

    The reaction

    \(A+B \rightleftharpoons X+Y\)

    has Kc = 1977 at 472K. At equilibrium _____.

    1. only products exist
    2. only reactants exist
    3. products predominate
    4. reactants predominate
    5. roughly equal molar amounts of products and reactants are present
    Answer

    c. products predominate

     

    Exercise \(\PageIndex{f}\)

    The equilibrium constant for reaction (1) is K. What is the equilibrium contant for equation (2)?

    (1)\(SO_{2}(g) + \frac{1}{2}O_{2} \rightleftharpoons SO_{3}(g)\)

    (2)\(2SO_{3}(g) \rightleftharpoons 2SO_{2}(g) + O_{2}(g)\)

    1. 1/2K
    2. 1/K2
    3. 2K
    4. K2
    5. -K2
    Answer

    b. 1/K2

     

    Exercise \(\PageIndex{g}\)

    The equilibrium constant for reaction (1) is K. What is the equilibrium constant for equation (2)?

    (1)\(\frac{1}{3}N_{2}(g) + H_{2}(g) \rightleftharpoons \frac{2}{3}NH_{3}(g)\)

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

    1. K3
    2. 3K
    3. K/3
    4. 1/K3
    5. -K3
    Answer

    d. 1/K3

     

    Exercise \(\PageIndex{h}\)

    If the value of Kc for the following reaction is 0.25:

    \(SO_{2}(g)+NO_{2}(g)\rightleftharpoons SO_{3}+NO(g)\)

    What is the value of Kc for the reaction below?

    \(2SO_{2}(g)+2NO_{2}(g)\rightleftharpoons 2SO_{3}+2NO(g)\)

    Answer

    \[K_{c1}=\frac{\left [ SO_{3} \right ]\left [ NO \right ]}{\left [ SO_{2} \right ]\left [ NO_{2} \right ]}=0.25\nonumber \]

    \[K_{c2}=\left ( K_{c1} \right )^{2}=\left (\frac{\left [ SO_{3} \right ]\left [ NO \right ]}{\left [ SO_{2} \right ]\left [ NO_{2} \right ]}\right )^{2}=0.25^{2}=0.0623 \nonumber \]

     

    Exercise \(\PageIndex{i}\)

    What is Kp for the following reaction at 25°C, Kc = 3.0*105?

    \(2H_{2}S(g)+3O_{2}\rightleftharpoons 2H_{2}O(g)+2SO_{2}(g)\)

    Answer

    \[T = 25+273.15 = 298.15\nonumber \]

    \[\Delta n = 4-5 = -1\nonumber \]

    \[K_{p}=K_{c}\left ( RT \right )^{\Delta n} \nonumber \]

    \[K_{p}=\left ( 3.0*10^{5} \right )\left ( 0.0821*298.15 \right )^{-1}=1.2*10^{4}\nonumber \]

     

    Exercise \(\PageIndex{j}\)

    The value of Kc for the reaction below is 2.0*10-1.0 at 100°C.

    \(COCl_{2}(g)\rightleftharpoons CO(g)+Cl_{2}(g)\)

    What is the value of Kc for the reverse reaction at 100°C?

    \(CO(g)+Cl_{2}(g)\rightleftharpoons COCl_{2}(g)\)

    Answer

    \[K_{c1}=2.0*10^{-10} \nonumber \]

    \[K_{c2}=\frac{1}{K_{c1}}=\frac{1}{2.0*10^{-10}}=5.0*10^{9}\]

     

    Exercise \(\PageIndex{k}\)

    The value of Kc for the following reaction is 1.10 at 25.0°C. What is the value of Kp for this reaction?

    \(4CuO(s)+CH_{4}(g) \rightleftharpoons CO_{2}(g) + 4Cu(s) + 2H_{2}O(g)\)

    Answer

    \[T = 25+273.15 = 298.15\nonumber \]

    \[\Delta n = 7-5 = 2 \nonumber \]

    \[K_{p}=K_{c}\left ( RT \right )^{\Delta n} \nonumber \]

    \[K_{p}=\left(1.10 \right )\left ( 0.0821*298.15 \right )^{2}=6.59*10^{2}\nonumber \]

     

    Exercise \(\PageIndex{l}\)

    What is the value of Kc for a flask at equilibrium that contains 0.0114 M HCl, 0.0931 M Cl2, and 0.0154 M H2 at a certain temperature?

    \(2HCl(g) \rightleftharpoons Cl_{2}(g) + H_{2}(g)\)

    Answer

    \[K_{c}=\frac{\left [ Cl_{2} \right ]\left [ H_{2} \right ]}{\left [ HCl \right ]^{2}}=\frac{\left [0.0931\right ]\left [0.0154\right ]}{\left [0.0114\right ]^{2}}=11.0\nonumber \]

     

    Exercise \(\PageIndex{m}\)

    Consider the gaseous equilibrium:\(2A\rightarrow 2B+C\)

    Determine the value of the missing B concentration at equilibrium.

    Exp #

    [A] at equilibrium

    [B] at equilibrium

    [C] at equilibrium

    1

    0.10 M

    0.10 M

    0.20 M

    2

    0.20 M

    0.50 M

    0.032 M

    3

    0.35 M

    ?

    0.15 M

    Answer

    Find k by solving the equilibrium constant expression using either experiment 1 or experiment 2 data

    \[k=\frac{\left [ B \right ]^{2}\left [ C \right ]}{\left [ A \right ]^{2}}=\frac{0.10^{2}*0.20}{0.10^{2}}=0.20\nonumber \]

    Solve the equilibrium constant expression for [B] then use k from above and solve with experiment 3 data

    \[\left [ B \right ]=\left ( \frac{k\left [ A \right ]^{2}}{\left [ C \right ]} \right )^{1/2}=\left ( \frac{0.20*\left [ 0.35 \right ]^{2}}{\left [ 0.15 \right ]} \right )^{1/2}=1.0 M\nonumber \]

     

    Exercise \(\PageIndex{n}\)

    Which of the following reactions at equilibrium has the following equilibrium constant expression?

    \(\frac{\left [IBr\right ]}{\left [I_{2}\right ]\left [Br_{2}\right ]}\)

    1. \(I_{2}(g)+Br_{2}(g)\rightleftharpoons IBr(g)\)
    2. \(I_{2}(g)+Br_{2}(g)\rightleftharpoons 2 IBr(g)\)
    3. \(2 IBr(g)\rightleftharpoons I_{2}(g)+Br_{2}(g)\)
    4. \(2 I_{2}(g)+2 Br_{2}(g)\rightleftharpoons IBr(g)\)
    5. \(IBr(g)\rightleftharpoons 2I_{2}(g)+2Br_{2}(g)\)
    Answer

    b. \(I_{2}(g)+Br_{2}(g)\rightleftharpoons 2 IBr(g)\)

     

    Exercise \(\PageIndex{o}\)

    The value of Kc for the following reaction at equilibrium is 54.0 at 427°C.

    \(H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)\)

    At this temperature, what is the value of Kc for:

    \(HI(g)\rightleftharpoons \frac{1}{2}H_{2}(g)+\frac{1}{2}I_{2}(g)\)

    Answer

    \[K_{c}^{1}=54.0 \nonumber \]

    \[K_{c}^{-1}=\frac{1}{\left (K_{c}^{1}\right )^{n}} \nonumber \]

    n is the factor of difference between the coefficients of the two reactions. So in this case it is 1/2.

    \[K_{c}^{-1}=\frac{1}{54.0^{1/2}}=\frac{1}{7.35}=0.136\nonumber \]

     

    Exercise \(\PageIndex{p}\)

    What is the equilibrium constant expression for the following reaction?

    \(Al_{2}\left (SO_{3}\right )_{3}(s)+6HCl(g)\rightleftharpoons 2AlCl_{3}(s)+3H_{2}O(l)+3SO_{2}(g) \)

    Answer

    \[K_{c}=\frac{\left [SO_{2}\right ]^{3}}{\left [HCl\right ]^{6}}\nonumber \]

     

    Exercise \(\PageIndex{q}\)

    What is the equilibrium constant expression for the following reaction?

    \(3SO_{2}(g)\rightleftharpoons 2SO_{3}(g)+S(s)\)

    Answer

    \[K_{c}=\frac{\left [SO_{3}\right ]^{2}}{\left [SO_{2}\right ]^{3}}\nonumber \]

     

    Exercise \(\PageIndex{r}\)

    Consider the following chemical reaction:

    \(H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)\)

    At equilibrium, the concentrations of H2, I2, and HI were 0.15M, 0.033 M, and 0.55 M, respectively. What is the Kc for this reaction?

    Answer

    \[K_{c}=\frac{\left [HI\right ]^{2}}{\left [H_{2}\right ]\left [I_{2}\right ]}=\frac{\left [0.55\right ]^{2}}{\left [0.15\right ]\left [0.033\right ]}=61\nonumber \]

     

    Exercise \(\PageIndex{s}\)

    \(PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)\)

    Initially, 1.26 mol of PCl5(g) was placed in a 1.0 L flask. At equilibrium, 1.08 mol of PCl5(g) was present. What is the value of Kc for this reaction at this temperature?

    Answer
    R PCl5 PCl3     + Cl2
    I 1.26   0 0
    C -x   +x +x
    E 1.08   x x

    \[1.08=1.26-x \nonumber\]

    \[x=1.26-1.08=0.18 \nonumber\]

    \[K_{c}=\frac{\left [ PCl_{3} \right ]\left [Cl_{2} \right ]}{\left [ PCl_{5} \right ]}=\frac{x^{2}}{1.26}=\frac{0.18^{2}}{1.26}=0.03\nonumber\]

     

    Exercise \(\PageIndex{t}\)

    \(PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g)\)

    What is the equilibrium partial pressure of PCl3? If in a 3.00 L vessel that was charged with 0.123 atm of PCl5 has a Kp of 0.0121?

    Answer
    R PCl5 PCl3     + Cl2
    I 0.123   0 0
    C -x   +x +x
    E 0.123-x   x x

    \[K_{p}=\frac{\left [PCl_{3}\right ]\left [Cl_{2}\right ]}{\left [PCl_{5}\right ]}=\frac{x^{2}}{0.123-x}=0.0121\nonumber\]

    \[\frac{x^{2}}{0.123-x}=0.0121 \nonumber\]

    \[x^{2}=\left (0.123-x\right )*0.0121 \nonumber\]

    \[x^{2}=0.001488-0.0121x \nonumber\]

    \[0=x^{2}+0.0121x-0.001488\nonumber\]

    Use the quadratic formual to solve for x

    \[x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-0.0121\pm \sqrt{0.0121^{2}-\left ( 4*1*-0.001488 \right )}}{2*1}\nonumber\]

    \[x=0.0330\nonumber\]

    The negative square is rejected because pressure cannot be negative.

     

    Exercise \(\PageIndex{u}\)

    The value of Kc for the following reaction is 0.070. What is the equilibrium concentration (M) of C4H10 if the equilibrium concentrations of C2H6 and C2H4 are both 0.035M?

    \(C_{4}H_{10}(g)\rightleftharpoons C_{2}H_{6}(g)+C_{2}H_{4}(g)\)

    Answer
    R C4H10 C2H6     + C2H4
    I x   0 0
    C -y   +y +y
    E x-y   y y

    \[\left [C_{2}H_{6}\right ]=\left [C_{2}H_{4}\right ]=0.035\,M=y \nonumber\]

    \[K_{c}=\frac{\left [C_{2}H_{6}\right ]\left [C_{2}H_{4}\right ]}{\left [C_{4}H_{10}\right ]}=\frac{0.035^{2}}{x-0.035}=0.070 \nonumber\]

    \[0.070x-0.00245=0.001225 \nonumber\]

    \[0.070x=0.003675 \nonumber\]

    \[x=0.0525 \nonumber\]

    \[\left [C_{4}H_{10}\right ]=x-y=0.0525-0.035=0.018M\nonumber\]

     

    Exercise \(\PageIndex{v}\)

    Nitrosyl bromide decomposes according to the following equation:

    \(2NOBr(g)\rightleftharpoons 2NO(g) + Br_{2}(g)\)

     

    A sample of NOBr (0.64 mol) was placed in a 1.00 L flask containing no NO or Br2. At equilibrium, the flask contained 0.46 mol of NOBr. How many moles of NO and Br2 are in the flask at equilibrium?

    Answer
    R 2NOBr 2NO     + Br2
    I 0.64 mol   0 0
    C -2x   +2x +x
    E 0.64-2x   2x x

    \[0.64-2x=0.46 \nonumber\]

    \[-2x=-0.18 \nonumber\]

    \[x=0.09 \nonumber\]

    \[mol_{NO}=2x=2*0.09=0.18\,mol \nonumber\]

    \[mol_{Br_{2}}=x=0.09\,mol \nonumber\]

     

    Exercise \(\PageIndex{w}\)

    Which of the following will shift to the left in response to a decrease in volume?

    1. \(H_{2}(g)+Cl_{2}(g)\rightleftharpoons 2HCl(g)\)
    2. \(N_{2}(g) + 3H_{2}(g)\rightleftharpoons 2NH_{3}(g)\)
    3. \(2SO_{3}(g)\rightleftharpoons 2SO_{2}(g)+O_{2}(g)\)
    4. \(2HI(g) \rightleftharpoons H_{2}(g)+I_{2}(g)\)
    5. \(4Fe(s)+3O_{2}(g)\rightleftharpoons 2Fe_{2}O_{3}(s)\)
    Answer

    c. \(2SO_{3}(g)\rightleftharpoons 2SO_{2}(g)+O_{2}(g)\)

     

    Exercise \(\PageIndex{x}\)

    For the endothermic reaction

    \(CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2}(g)\)

     

    only _____ would favor shifting the equilibrium position to form more CO2 gas.

    1. both decreasing the system temperature and increasing the system pressure
    2. decreasing the system temperature
    3. increasing both the system temperature and the system pressure
    4. increasing the system pressure
    5. increasing the system temperature
    Answer

    e. increasing the system temperature

     

    Exercise \(\PageIndex{y}\)

    Which of the following reactions would increase pressure at constant temperature not change the concentration of reactants and products?

    1. \(2N_{2}(g)+O_{2}(g)\rightleftharpoons 2N_{2}O(g)\)
    2. \(N_{2}(g)+2O_{2}(g)\rightleftharpoons 2NO_{2}(g)\)
    3. \(N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)\)
    4. \(N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)\)
    5. \(N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)\)
    Answer

    e. \(N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)\)

     

    Exercise \(\PageIndex{z}\)

    Consider the following reaction at equilibrium

    \(2CO_{2}\rightleftharpoons 2CO(g)+O_{2}(g)\)

     

    The yield of CO(g) in reaction can be maximized by carrying out the reaction _____.

    1. at high temperature and high pressure
    2. at high temperature and low pressure
    3. at low temperature and high pressure
    4. at low temperature and low pressure
    5. in the presence of solid carbon
    Answer

    d. at low temperature and low pressure

     

    Exercise \(\PageIndex{aa}\)

    The effect of a catalyst on a chemical reaction is to _____.

    1. accelerate the forward reaction only
    2. increase the entropy change associated with a reaction
    3. lower the energy of the transition state
    4. make reactions more exothermic
    5. react with product, effectively removing it and shifting the equilibrium to the right
    Answer

    c. lower the energy of the transition state

    Textbook: 15: Equilibria

    Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

    • Liliane Poirot