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For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction N2(g) + 3H2(g) -----> 2NH(g).the standard change in Gibbs free energy is ΔG° = -32.8 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are. Pn2= 0.1 atm, Ph2=.1 atm and P (NH3)=.6 atm. delta g=?

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jufsanabriasa

Answer:

Explanation:

It is possible to obtain ΔG of a reaction using:

ΔG = ΔG° + RT ln Q

Where:

ΔG° is standard change in Gibbs free energy (-32,8 kJ/mol)

R is gas constant (8,314472x10⁻³ kJ/molK)

T is temperature (298 K)

And Q is reaction quotient. For the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Q is: [NH₃]² /[H₂]³[N₂]

Replacing:

ΔG = -32,8kJ/mol + 8,314472x10⁻³ kJ/molK×298 K ln [0,6]² /[0,1]³[0,1]

ΔG = -12,5 kJ/mol

I hope it helps!

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