Question

The standard Gibbs energy change for the reaction
NH3(aq)+H2O(l)?NH+4(aq)+OH?(aq)
is 29.05 kJmol?1 at 298 K.

Use this thermodynamic quantity to decide in which direction the reaction is spontaneous when the concentrations of NH3(aq), NH+4(aq), and OH?(aq) are 0.10 M, 1.0

Answer 1

Given :

Delta G = 29.05 kJ/mol = 29050 J/mol

T = 298 K

[NH3]= 0.10 M, [NH4+] = 1.0 X 10^-3 M , [OH-]= 1.0 X 10^-3 M

We calculate equilibrium constant of the reaction by using Delta G

The relation between Delta G and K is as follow :

Delta G = -RTlnK

Delta G is Gibbs free energy change, R is gas constant = 8.314 J/Kmol ,T is temperature

Lets calculate K by using given value of T and Delta G

Ln K = -Delta G / RT

=-(29050 J/mol ) / [(8.314 J /K mol )*298 K]

=-11.7252

ln K = -11.7252

Lets take exp of both side

K = exp (-11.7252)

= 8.08752 X 10^-6

Now we got K

From given concentration date we calculate Q

Q = ( [NH+4(aq)][OH-(aq)])/ [NH3(aq)]

(H2O doesnot involve )

Lets plug given value in above Q formula

= (1.0 X 10^-3 M)²/ 0.1

= 1.00 X 10^-5

The condition for forward reaction

Q < K

Condition for reverse reaction

Q>K

Condition for equilibrium

Q=K

Now we got Q = 1.00 X 10^-5

And K = 8.08752 X 10^-6

So Q is greater than K

And reaction is spontaneous in the reverse direction.