RELATION BETWEEN KP AND KC

For a general equation,

                              aA + bB  1573_equal.JPG cC + dD

Where, a , b, c and d are coefficients of the reacting substance

                              Kp = (PC)c (OD)d / (PA)a (PB)b

From gas equation 

                            PV = nRT        =>  P = n/V RT

So,                       PA = [A]RT’ PB = [B]RT 

                            PC = [C]RT; PD = [D]RT

Hence,           Kp = ([C]RT)c.([D]RT)d / ([A]RT)a / ([B]RT)b

                     or, Kp = [C]d[D]d / [A]a[B]b × (RT)[(c+d) – (a+b)] 

                     Where Δng = {(c + d) – (a + b)} = change in the numbers of gaseous moles.

Hence,           Kp = Kc.(RT)Δng

                     When       Δn = 0; Kp = Kc

                    Δn > 0; Kp > Kc

                    Δn < 0;            Kp < Kc

Note: Similarly we can find equilibrium constant (Kx) interms of mole fraction and can find out its relation with Kp and Kc.                  

Illustration . At 27°C Kp value for reaction  is 0.1 atm, calculate its Kc value.            

Solution:       K= Kc(RT)Dn

                        Δn = 1

                        Kc = KP / RT = 0.1 / 0.82 × 300 = 4 × 10-3

Illustration . Solid NH4I dissociates according to the reaction at 400 K

                   NH4I(s) 1068_equal.JPG NH3(g) + HI(g) ; Kp = 16 atm. In presence of catalyst HI dissociates in H2 and I2 as 2HI 1068_equal.JPG H2 + I2. If partial pressure of H2 at this temp is 1 atm in the container when both the equilibrium exist simultaneously, calculate Kp value of second equilibrium (for the dissociation of HI).                                                                                                        

Solution:      NH4I(s) 1068_equal.JPG NH(g) + HI (g)            KP1 = 16 

                        2HI 1068_equal.JPG H2 +     I2

                        P - x            x/2 . x/2                    KP2 = ?

                        Here x/2 = 1 atm

                        x = 2 atm

                        P(P -2) = 16

                        P = 5.1

Illustration . A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on addition of graphite. What is the value of Kp if the total pressure at equilibrium is 0.8 atm?        

Solution: CO2(g)      +    C(s)     ——>    2CO(g)

                         0.5                                    2n                                                                                        

                        or, 0.5-n                               0.5+n = 0.8 

                        n = 0.3

                        Kp = (2×0.3)2 / (0.5–0.3) = 1.8 atm

Illustration . Given that the equilibrium constant for the reaction, H2(g) + I2(g) 1068_equal.JPG 2HI(g) is 50 at 700K. Calculate the equilibrium constant for the reaction.

                        HI(g) 1068_equal.JPG 1/2 H2(g) + 1/2 l2(g)

Solution: We have H2(g) + I2(g) 1068_equal.JPG 2HI(g)

                        Kc = [Hl]2 / [H2][l2] = 50

                        The new equilibria is

                        HI(g) 1068_equal.JPG [H2]1/2[l2]1/2 / [Hl] = 1/√Kc

                        = 1/ √50 = √2 / 10 = 0.141

Illustration . One mole of N2 is mixed with 3 moles of H2 in a 4 litre container. If 25% of N2 is converted into NH3 by the following reaction

                   N2(g) + 3H2(g) 1068_equal.JPG 2NH3(g). Calculate Kc and Kp of the reaction. (Temperature = 227°C and R = 0.08231).

Solution: We have N2(g) + 3H2(g) 1068_equal.JPG 2NH3(g)

                        Percentage N2 reacted, 25%

                        x = 0.25

                        Now, (a – x) = 1 – 0.25 = 0.75

                        b – 3x = 3 – 0.75 = 2.25

 2003_equation.JPG

                        = 1.48 × 10–5 L2 mol–2

                        Now, Kp = Kc . (RT)Δn = 1.48 × 10–5 × [0.0821] × (227 + 273)–2

                        1.48 × 10–5 / {0.082 × (500)}2

                            = 8.78 × 10–9

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