Chemical Equilibrium : Reaction quotient Kc Kp

 

Applications of equilibrium constant : Knowing the value of the equilibrium constant for a chemical reaction is important in many ways. For example, it judge the extent of the reaction and predict the direction of the reaction.

Judging the extent of reaction

We can make the following generalisations concerning the composition of equilibrium mixture.

(a)     If Kc > 103, products predominate over reactants. If Kc is very large, the reaction proceeds almost all the way to completion.

(b)     If Kc < 10–3, reactants predominate over products. If Kc is very small, the reaction proceeds hardly at all.

(c)      If Kc is in the range 10–3 to 103, appreaciable concentration of both reactants and products are present. This is illustrated as follows,

 

Reaction quotient and predicting the direction of reaction : The concentration ratio, i.e., ratio of the product of concentrations of products to that of reactants is also known as concentration quotient and is denoted by Q.

Concentration quotient, Q = \frac { [X][Y] }{ [A][B] }

It may be noted that Q becomes equal to equilibrium constant (K) when the reaction is at the equilibrium state. At equilibrium, Q = K = Kc = Kp. Thus,

(a)     If Q > K, the reaction will proceed in the direction of reactants (reverse reaction).

(b)     If Q < K, the reaction will proceed in the direction of the products (forward reaction).

(c)      If Q = K, the reaction mixture is already at equilibrium.

Thus, a reaction has a tendency to form products if Q < K and to form reactants if Q > K.

This has also been shown in figure,

Calculation of equilibrium constant : We have studied in the characteristics of the equilibrium constant that its value does not depend upon the original concentrations of the reactants and products involved in the reaction. However, its value depends upon their concentrations at the equilibrium point. Thus, if the equilibrium concentrations of the species taking part in the reaction be known, then the value of the equilibrium constant and vice versa can be calculated.

 

Homogeneous equilibria and equations for equilibrium constant

(Equilibrium pressure is P atm in a V L flask)

Δn = 0; Kp = Kc Δn < 0; Kp < Kc Δn > 0; Kp < Kc
\[\underset{(g)}{\mathop{{{H}_{2}}}}\,+\underset{(g)}{\mathop{{{I}_{2}}}}\,\rightleftharpoons \underset{(g)}{\mathop{2HI}}\,  \[\underset{(g)}{\mathop{{{N}_{2}}}}\,+\underset{(g)}{\mathop{3{{H}_{2}}}}\,\rightleftharpoons \underset{(g)}{\mathop{2N{{H}_{3}}}}\, \[\underset{(g)}{\mathop{2S{{O}_{2}}}}\,+\underset{(g)}{\mathop{{{O}_{2}}}}\,\rightleftharpoons \underset{(g)}{\mathop{2S{{O}_{3}}}}\, \[\underset{(g)}{\mathop{PC{{l}_{5}}}}\,\rightleftharpoons \underset{(g)}{\mathop{PC{{l}_{3}}}}\,+\underset{(g)}{\mathop{C{{l}_{2}}}}\,
Initial mole   1        1            0 1          3         0    2           1            0 1           0         0
Mole at Equilibrium (1–x) (1– x)    2x (1–x) (3–3x)   2x (2–2x) (1–x)        2x (1–x)           x       x
Total mole at equilibrium 2 (4 – 2x) (3 – x) (1 + x)
Active masses \[\left( \frac{1-x}{V} \right)  \[\left( \frac{1-x}{V} \right)  \[\frac{2x}{V}

    \[\left( \frac{1-x}{V} \right)\] \[3\,\,\left( \frac{1-x}{V} \right)\] \[\left( \frac{2x}{V} \right)

    \[\left( \frac{2-2x}{V} \right)\]\[\left( \frac{1-x}{V} \right)\] \[\left( \frac{2x}{V} \right)

    \[\left( \frac{1-x}{V} \right)\]\[\left( \frac{x}{V} \right)\] \[\left( \frac{x}{V} \right)
Mole fraction

    \[\left( \frac{1-x}{2} \right)\]\[\left( \frac{1-x}{2} \right)\] \[\frac{2x}{2}

   

    \[\frac{1-x}{2\,\,\,\left( 2-x \right)}\]\[\frac{3}{2}\left( \frac{1-x}{2-x} \right)\]\[\frac{x}{(2-x)}

 

    \[\left( \frac{2-2x}{3-x} \right)\] \[\left( \frac{1-x}{3-x} \right)\,\,\,\left( \frac{2x}{3-x} \right)

    

    \[\left( \frac{1-x}{1+x} \right)\] \[\left( \frac{x}{1+x} \right)\] \[\left( \frac{x}{1+x} \right)

    
Partial pressure

    \[p\,\,\,\left( \frac{1-x}{2} \right)\]\[p\,\,\,\left( \frac{1-x}{2} \right)\]\[p\,\,\,\left( \frac{2x}{2} \right)

    \[P\,\,\left( \frac{1-x}{2\,\,(2-x)} \right)\stackrel{\scriptscriptstyle\to}{\leftarrow}\]\[\left( \frac{3(1-x)}{2(2-x)} \right)\,\,P\,\,\frac{Px}{(2-x)}

    \[P\,\,\,\left( \frac{2-2x}{3-x} \right)\] \[P\,\,\,\left( \frac{1-x}{3-x} \right)\] \[P\,\,\,\left( \frac{2x}{3-x} \right)

    \[P\,\,\,\left( \frac{1-x}{1+x} \right)\] \[P\,\,\,\left( \frac{x}{1+x} \right)\] \[P\,\,\,\left( \frac{x}{1+x} \right)
Kc \[\frac{4{{x}^{2}}}{{{\left( 1-x \right)}^{2}}} \[\frac{4{{x}^{2}}{{V}^{2}}}{27\,\,{{\left( 1-x \right)}^{4}}} \[\frac{{{x}^{2}}V}{{{\left( 1-x \right)}^{3}}} \[\frac{{{x}^{2}}}{\left( 1-x \right)V}
Kp  \[\frac{4{{x}^{2}}}{{{\left( 1-x \right)}^{2}}} \[\frac{16{{x}^{2}}\,\,{{\left( 2-x \right)}^{2}}}{27\,\,{{\left( 1-x \right)}^{4}}{{P}^{2}}} \[\frac{P{{x}^{2}}}{\left( 1-{{x}^{2}} \right)} \[\frac{P{{x}^{2}}}{\left( 1-{{x}^{2}} \right)\stackrel{\scriptscriptstyle\to}{\leftarrow}}

 

Heterogeneous equilibria and equation for equilibrium constant

(Equilibrium pressure is P atm)

NH4HS(s) ⇌ NH3(g) + H2S(g) C(s) + CO2(g) ⇌ 2CO(g) NH2CO2NH4(s) ⇌ 2NH3(g) + CO2(g)
Initial mole 1     0        0 1      1       0 1          0     0
Mole at equilibrium (1–x) x     x (1–x) (1–x) 2x (1–x)   2x     x
Total moles at equilibrium (solid not included) 2x   (1+x) 3x
Mole fraction \[\frac{x}{2x}=\frac{1}{2}\,\,\,\,\frac{1}{2}     

    \[\left( \frac{1-x}{1+x} \right)\] \[\left( \frac{2x}{1+x} \right)

    \[\frac{2}{3}\] \[\frac{1}{3}

       
Partial pressure \[\frac{P}{2}\,\,\,\frac{P}{2}     \[P\left( \frac{1-x}{1+x} \right)\,\,\,\,P\left( \frac{2x}{1+x} \right) \[\frac{2P}{3}\,\,\,\,\frac{P}{3}        
Kp \[\frac{{{P}^{2}}}{4}  \[\frac{4{{P}^{2}}{{x}^{2}}}{(1-{{x}^{2}})}  \[\frac{4{{P}^{3}}}{27}

 

Equilibrium constant and standard free energy change : Standard free energy change of a reaction and its equilibrium constant are related to each other at temperature T by the following relation,

                                      ΔGo = –2.303 RT log K

when, ΔHo = – ve, the value of equilibrium constant will be large positive quantity and

when, ΔGo = + ve, the value of equilibrium constant is less than 1 i.e., low concentration of products at equilibrium state.

 

Examples based on Equilibrium constant, Kp & Kc and ΔGo = –2.303 RT log K

Example 1 :      In the reversible reaction A + B ⇌ C + D, the concentration of each C and D at equilibrium was 0.8 mole/litre, then the equilibrium constant Kf will be

  (a) 6.4                            (b) 0.64                (c) 1.6                             (d) 16.0

Solution: (d)     Suppose 1 mole of A and B each taken then 0.8 mole/litre of C and D each formed remaining concentration of A and B will be (1 – 0.8) = 0.2 mole/litre each.

Kc = \frac { [C][D] }{ [A][B] } = \frac { 0.8\times 0.8 }{ 0.2\times 0.2 } = 16.0

 

Example 2 :      For the system A(g) + 2B(g) ⇌ C(g), the equilibrium concentrations are (A) 0.06 mole/litre (B) 0.12 mole/litre (C)0.216 mole/litre. The Keq for the reaction is

(a) 250                           (b) 416                 (c) 4 × 10–3                     (d) 125

Solution: (a)     For reaction A + 2B ⇌ C ;

Keq = \frac { [C] }{ [A][B]^{ 2 } } = \frac { 0.216 }{ 0.06\times 0.12\times 0.12 } = 250  

 

Example 3 :      Molar concentration of O2 is 96 gm, it contained in 2 litre vessel, active mass will be

(a) 16 mole/litre                               (b) 1.5 mole/litre

(c) 4 mole/litre                                 (d) 24 mole/litre

Solution: (b)     Active mass \frac { \frac { weight }{ M.wt } }{ Volume } =\frac { weight }{ M.wt\times Volume } =\frac { 96 }{ 32\times 2 } \frac { 3 }{ 2 } = 1.5 mol/litre

 

Example 4 :      2 moles of PCl5 were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5 is dissociated into PCl3 and Cl2. The value of equilibrium constant is

  (a) 0.266                        (b) 0.53                (c) 2.66                           (d) 5.3

Solution: (a)               At start,  PCl5 ⇌  PCl3 + Cl2

                                      At equilibrium.  \frac { 2\times 60 }{ 100 } \frac { 2\times 40 }{ 100 } \frac { 2\times 40 }{ 100 }  

                                      Volume of container = 2 litre

                                      Kc = \frac { \frac { 2\times 40 }{ 100\times 2 } \times \frac { 2\times 40 }{ 100\times 2 } }{ \frac { 2\times 60 }{ 100\times 2 } }   = 0.266

 

Example 5 :      A mixture of 0.3 mole of H2 and 0.3 mole of  is allowed to react in a 10 litre evacuated flask at 500oC. The reaction is H2 + I2   ⇌ 2HI the Kc is found to be 64. The amount of unreacted I2 at equilibrium is

  (a) 0.15 mole                (b) 0.06 mole      (c) 0.03 mole                 (d) 0.2 mole

Solution: (b)              Kc = \frac { { [HI] }^{ 2 } }{ { [H }_{ 2 }]{ [I }_{ 2 }] }   ;  64 = \frac { { x }^{ 2 } }{ 0.03\times 0.03 }    

                                      x2 = 64 × 9 × 10–4 ; x = 8 × 3 × 10–2 = 0.24

x is the amount of HI at equilibrium. Amount of I2 at equilibrium will be 0.30 – 0.24 = 0.06 mole

 

Example 6 :      The rate constant for forward and backward reactions of hydrolysis of ester are 1.1 × 10–2 and 1.5 × 10–3 per minute respectively. Equilibrium constant for the reaction,

CH3COOC2H5 + H2O ⇌ CH3COOH + C2H5OH is

  (a) 4.33                          (b) 5.33                (c) 6.33                           (d) 7.33

Solution: (d)               Kf = 1.1 × 10–2, Kb = 1.5 × 10–3 ;

         Kc = \frac { { K }_{ f } }{ { K }_{ b } } =\frac { { 1.1 }\times { 10 }^{ -2 } }{ 1.5\times { 10 }^{ -3 } }   = 7.33

 

Example 7 :      For the reaction PCl3(g) + Cl2(g) ⇌ PCl5 at , the value of Kc is 26, then the value of Kp on the same temperature will be

  (a) 0.61                          (b) 0.57                (c) 0.83                           (d) 0.46

Solution: (a)               Dng = 1 – 2 = –1

                                      Kp = Kc(RT)Δn;           Kp = Kc(RT)–1

Since R = 0.0821 litre atm k–1 mol–1, T = 250oC = 250 + 273 = 523 Kp = 26(0.0821 × 523)–1 = 0.61

 

Example 8 :      If Kp for reaction A(g) + 2B(g) ⇌ 3C(g) + D(g) is 0.05 atm at 1000 K its Kc in term of R will be

(a) \frac { 5\times { 10 }^{ -4 } }{ R }                    (b) \frac { 5 }{ R }                   (c) \frac { 5\times { 10 }^{ -5 } }{ R }          (d) None of these

Solution: (c)      Kp = Kc(RT)Δn   5 × 10–2 = Kc (R × 1000)1 ⇒ Kc = \frac { 5\times { 10 }^{ -5 } }{ R }

 

Example 9 :      If the equilibrium constant of the reaction  2HI ⇌ H2 + I2 is 0.25, then the equilibrium constant of the reaction  H2 + I2   ⇌ 2HI would be

  (a) 1.0                            (b) 2.0                  (c) 3.0                            (d) 4.0

Solution: (d)      for the IInd reaction is reverse of Ist for reaction 2HI ⇌ H2 + I2 is 0.25Kc for  reaction, H2 + I2 ⇌ 2HI will be K’ = \frac { 1 }{ { K }_{ c } } =\frac { 1 }{ 0.25 }   = 4.

 

Example 10 :    If equilibrium constant for reaction 2AB ⇌ A2 + B2, is 49, then the equilibrium constant for reaction   AB ⇌  \frac { 1 }{ 0.25 } A2 + \frac { 1 }{ 0.25 } B2, will be

  (a) 7                               (b) 20                   (c) 49                              (d) 21

Solution: (a)                 2AB ⇌ A2 + B2

                                       Kc =  \frac { { [A }_{ 2 }][B_{ 2 }] }{ { [AB] }^{ 2 } } = 49

                                        For reaction AB ⇌  \frac { 1 }{ 2 } A2 + \frac { 1 }{ 2 } B2

                                       Kc = \frac { { [A }_{ 2 }]^{ 1/2 }[B_{ 2 }]^{ 1/2 } }{ { [AB] } }

                                       K’c = \sqrt { { K }_{ c } } =\sqrt { 49 } =7

 

Example 11 :      For the reaction 2NO2(g) ⇌ 2NO(g) + O2(g) Kc = 1.8 × 10–6  at 185oC. At 185oC, the value of Kc for the reaction NO(g) + \frac { 1 }{ 2 } O2(g) ⇌ NO2(g) is

  (a) 0.9 × 106                  (b) 7.5 × 102        (c) 1.95 × 10–3     (d) 1.95 × 103

Solution: (b)       Reaction is reversed and halved.

                              \[\therefore \,\,\,\,\,{{K}_{c}}=\frac{1}{\sqrt{{{K}_{c}}}}  ;  \[{{{K}'}_{c}}=\frac{1}{\sqrt{1.8\times {{10}^{-6}}}}  = 7.5 × 102