Question

Nitrogen is a vital element for all living systems, but except for a few types of bacteria, blue-green algae, and some soil fungi, most organisms cannot utilize N2 from the atmosphere. The formation of "fixed" nitrogen is therefore necessary to sustain life, and the simplest form of fixed nitrogen is ammonia NH3. A possible pathway for ammonia synthesis by a living system is: 12N2(g)+32H2O(l)→NH3(aq)+32O2(g)

where (aq) means the ammonia is dissolved in water.

Substance and deltaG (formation):

N2(g) 0

H2O(l) -237.1

NH3(aq) -26.5

O2(g) 0

Delta Gf for the biological synthesis was 3.292*10^5 J*mol^-1 (not sure how to get this answer)

Question: Calculate the equilibrium constant, Kp, for the biological synthesis of ammonia?

Answer 1

The balanced equation for biological synthesis of one mole ammonia is

(1/2)N₂(g)+(3/2)H₂O(l)→NH₃(aq)+(3/4)O₂(g).

Now change in Gibbes energy for overall reaction,

_rG⁰ = (1)_fG⁰[NH₃(aq)] + (3/4)_fG⁰[O₂(g)] - (1/2)_fG⁰[N₂(g)] - (3/2)_fG⁰[H₂O(l)]

= (1)(- 26.5) + (3/4)(0) - (1/2)(0) - (3/2)(- 237.1) kJ.mol^-1

= 329.15 kJ.mol^-1

We know that K_P = - _rG⁰ / RT

= - ( 329.15*10^3) J.mol^-1 / (8.314 J.K^-1.mol^-1)*(298.15K) (considering standard temperature)

  = - 132.85

therefore K_p = 2.01*10^{^-58}