We define the activity of pure liquids as 1 to define the standard state. Howeve

Question

We define the activity of pure liquids as 1 to define the standard state. However, this is an arbitrary choice. Find the dissociation constant, K, for the autoprotolysis of pure water using the molarity standard state for water (i.e. defining the activity as 1 for a water concentration of 1 mol/l). You may take the density of water to be 1.0 g/ml.

Express your answer using three significant figures.

Part C

Estimate K in this standard state at 60?C. Assume constant enthalpy of reaction. ?rH?=55.82 kJ mol?1

Express your answer using three significant figures.

Explanation / Answer

Sol:-

(1) Autoprotolysis of water is :-

H2O(l) + H2O(l) <------------> H3O+(aq) + OH-(aq)

The expression of equilibrium constant (keq) of this reaction is :

keq = [H3O+] [OH-] / [H2O]2 ..........(1)

also ionisation of water is :-

H2O(l) <-----------> H+(aq) + OH-(aq)

Now the expression of dissociation constant kd is :

kd = [H+] [OH-] /[H2O] ...........(2)

at 250C [ H+ ] = [ OH- ] = 1.0 x 10-7 M and [ H2O ] = 55.55 M

Now substituting these values in equation (2) , we have

kd = (1.0 x 10-7) (1.0 x 10-7) / 55.55

kd = 1.0 x 10-14 / 55.55

kd = 1.8 x 10-16

Hence k of pure water = 1.80 x 10-16

(2)

From the mathematical expression of Vant Hoff equation i.e

log k2/k1 = detlaH / 2.303 R (1 / T1 - 1 / T2 )  

OR

log k2 - log k1 = detlaH / 2.303 R (T2 - T1 / T1 T2) ..........(3)

here k1 is the equilibrium constant at temperature T1 and k2 is the equilibrium constant at temperature T2 .

delta H = change in the enthalpy of the reaction = 55.82 KJ/mol .

k1 = 1.8 x 10-16

T1 = 250C = 298 K

k2 = ?

T2 = 600C = 333 K and

R = gas constant = 8.314 x 10-3 KJ K-1 mol-1

now from equation (3) , we have

log k2 - log 1.8 x 10-16 = 55.82 KJ/mol / 2.303 x 8.314 x 10-3 KJ K-1 mol-1 ( 333 K - 298 K / 333 K x 298 K )

log k2 - (- 15.74 ) = 2.91 x 103 ( 35 / 99234)

log k2 + 15.74 = 1.02 x 10-3 x 103

log k2 = 1.02 - 15.74

log k2 = - 14.72

k2 = 10-14.72

k2 = 1.90 x 10-15

Hence k at 600C is 1.90 x 10-15

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