Question

Nitrogen is a vital component of proteins and nucleic acids, and thus is necessary for life. The atmosphere is composed of roughly 80% N2, but most organisms cannot directly utilize N2 for biosynthesis. Bacteria capable of "fixing" nitrogen ( ie converting N2 to a chemical form, such as NH3, which can be utilized in the biosynthesis of proteins and nucleic acids) are called diazatrophs. The ability of some plants like legumes to fix nitrogen is due to a symbiotic relationship between the plant and nitrogen-fixing bacteria that live in the plant’s roots.

Assume the hypothetical reaction for fixing nitrogen biologically is N2 (g) + 3H2O (l) → 2 NH3 (aq) + 3/2 O2 (g)

a. Calculate the standard enthalpy change for the biosynthetic fixation of nitrogen at T = 298 K. For NH3 (aq), ammonia dissolved in aqueous solution, ΔHof = - 80.3 kJ mol-1.

b. In some bacteria, glycine is produced from ammonia by the reaction NH3 (aq) + 2CH4 (g) + 5/2 O2 (g) → NH2CH2COOH (s) + 3H2O (l) Calculate the standard enthalpy change for the synthesis of glycine from ammonia. For glycine, ΔHof = - 537.2 kJ mol -1. Assume that T = 298 K.

c. Calculate the standard enthalpy change for the synthesis of glycine from nitrogen, oxygen, and methane.

Answer 1

a. ΔHr = 696,8 kJ/mol

b. ΔHr = -1164,7 kJ/mol

c. ΔHr = -467,9 kJ/mol

Step-by-step explanation:

It is possible to obtain the standard enthalpy change of a reaction with the ΔH°f of products - ΔH°f reactants.

a. For the reaction:

N₂(g) + 3H₂O(l) → 2NH₃(aq) + ³/₂O₂(g)

ΔHr = 2ΔH°f NH₃(aq) + ³/₂ΔH°fO₂(g) - (3ΔH°fH₂O(l) + 2ΔH°f N₂(g))

ΔHr = 2×-80,3 kJ/mol + ³/₂×0 - (3×-285,8 kJ/mol + 0)

ΔHr = 696,8 kJ/mol

b. and c. For the reaction:

NH₃(aq) + 2CH₄(g) + ⁵/₂O₂(g) → NH₂CH₂COOH(s) + 3H₂O(l)

ΔHr = ΔH°fNH₂CH₂COOH(s) + 3ΔH°fH₂O(l) - (ΔH°fNH₃(aq) + 2ΔH°fCH₄(g) + ⁵/₂ΔH°fO₂(g))

ΔHr = -537,2kJ/mol + 3×-285,8 kJ/mol - (-80,3 kJ/mol + 2×-74,8kJ/mol+ ⁵/₂×0)

ΔHr = -1164,7 kJ/mol

c. From nitrogen, methane and oxygen the reaction is the sum of reactions of a and b:

N₂(g) + 3H₂O(l) → 2NH₃(aq) + ³/₂O₂(g)

+ NH₃(aq) + 2CH₄(g) + ⁵/₂O₂(g) → NH₂CH₂COOH(s) + 3H₂O(l)

N₂(g) + 2CH₄(g) + O₂(g) → NH₂CH₂COOH(s) + NH₃(aq)

By Hess's law, the ΔHr will be the sum of the ΔHr of the last two reactions, that means:

ΔHr = -1164,7 kJ/mol + 696.8 kJ/mol

ΔHr = -467,9 kJ/mol

I hope it helps!

Answer 2

Explanation:ANd of course,

Δ

H

∘

f

=

0

for an element (here dixoygen) in its standard state