Question

# 1. Nitrogen is a vital component of proteins and nucleic acids, and thus is necessary for...

1. 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.

Given reaction is N2 (g) + 3H2O (l) → 2 NH3 (aq) + 3/2 O2 (g)
Since you can see here, only NH3 and H2O is in liquid state, so Hf will be zero for O2 and N2
So , delta H for reaction = 2*(-80.3) - 3(-285.83) = 696.89 kJ/mol
Delta U = delta H - delta(n)*RT = 696.89-1/2*8.314*298*0.0001 (since converting to kJ will make 0.0001 multiplier and delta (n) for reaction is 1/2)
= 695.651 kJ/mol

b) standard enthalpy change for the synthesis of glycine from ammonia
Given equation is NH3 (aq) + 2CH4 (g) + 5/2 O2 (g) → NH2CH2COOH (s) + 3H2O (l)
Taking into consideration of number of molecules.
we have
delta H = 3*(-285.83) + (-537.2) -2*(-74.85) - (-45.90)
= -1199.09 kJ/mol
c) Given equation is 1/2 N2(g) + 7/2 O2(g) + 4CH4 → 2NH2CH2COOH + 3H2O (l)
delta H = 3*(-885.83) +2*(-537.2)-4*(-74.85) = -816.25 kJ/mol

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