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 ΔG∘f(kJ⋅mol−1)N2(g) 0H2O(l) −237.1NH3(aq) −26.5O2(g) 0Calculate ΔG∘f for the biological synthesis of ammonia at 298 K.Calculate the equilibrium constant for the biological synthesis of ammonia.

Answer 1

(a) The value of
`\Delta G^o_(rxn)` for the biological synthesis of ammonia at 298 K is
`3.29* 10^5J/mol`

(b) The value of equilibrium constant for the biological synthesis of ammonia is
`2.14* 10^(-58)`

Explanation :

(a) The equation used to calculate
`\Delta G^o_(rxn)` of a reaction is:

`\Delta G^o_(rxn)=\sum [n* \Delta G^o_f(product)]-\sum [n* \Delta G^o_f(reactant)]`

The equilibrium reaction follows:

`(1)/(2)N_2(g)+(3)/(2)H_2O(l)\rightleftharpoons NH_3(aq)+(3)/(2)O_2(g)`

The equation for the
`\Delta G^o_(rxn)` of the above reaction is:

`\Delta G^o_(rxn)=[(n_((NH_3))* \Delta G^o_f_((NH_3)))+(n_((O_2))* \Delta G^o_f_((O_2)))]-[(n_((N_2))* \Delta G^o_f_((N_2)))+(n_((H_2O))* \Delta G^o_f_((H_2O)))]`

We are given:

`\Delta G^o_f_((NH_3(aq)))=-26.5kJ/mol\\\Delta G^o_f_((O_2(g)))=0kJ/mol\\\Delta G^o_f_((N_2(g)))=0kJ/mol\\\Delta G^o_f_((H_2O(l)))=-237.1kJ/mol`

Putting values in above equation, we get:

`\Delta G^o_(rxn)=[(1* -26.5)+((3)/(2)* -0)]-[((1)/(2)* 0)+((3)/(2)* -237.1)]=329.15kJ/mol=3.29* 10^5J/mol`

Thus, the value of
`\Delta G^o_(rxn)` for the biological synthesis of ammonia at 298 K is
`3.29* 10^5J/mol`

(b) Now we have to calculate the value of equilibrium constant.

Formula used :

`\Delta G^o=-RT* \ln K`

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in this formula, we get the value of K.

`3.29* 10^5J/mol=-(8.314 J/K/mole)* (298K)* \ln K`

`K=2.14* 10^(-58)`

Therefore, the value of equilibrium constant for the biological synthesis of ammonia is
`2.14* 10^(-58)`