Ammonia can be produced by the reaction of hydrogen gas and nitrogen gas, as shown below:N2(g) + 3H2(g) → 2NH3(g)Given that the standard free energy of formation (∆Gof) of NH3 (g) is -105 kJ/mol at 304 K, calculate the equilibrium constant, K, at this temperature.To express an answer in exponential notation, use E to indicate the exponent. For example, 3.0 x 103 would be written, 3.0E3.K = Ammonia can be produced by the reaction of hydrogen gas and nitrogen gas, as shown below:N₂(g) + 3H₂(g) → 2NH3(g)Given that the standard free energy of formation (AG°f) of NH3 (g) is -105 kJ/mol at 304 K, calculate the equilibrium constant, K, at this temperature.To express an answer in exponential notation, use E to indicate the exponent. For example, 3.0 x 10³ would be written, 3.0E3.K=

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Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 15P

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Ammonia can be produced by the reaction of hydrogen gas and nitrogen gas, as shown below:

N₂(g) + 3H₂(g) → 2NH₃(g)

Given that the standard free energy of formation (∆G^o_f) of NH₃ (g) is -105 kJ/mol at 304 K, calculate the equilibrium constant, K, at this temperature.
To express an answer in exponential notation, use E to indicate the exponent. For example, 3.0 x 10³ would be written, 3.0E3.

K =

Ammonia can be produced by the reaction of hydrogen gas and nitrogen gas, as shown below:
N₂(g) + 3H₂(g) → 2NH3(g)
Given that the standard free energy of formation (AG°f) of NH3 (g) is -105 kJ/mol at 304 K, calculate the equilibrium constant, K, at this temperature.
To express an answer in exponential notation, use E to indicate the exponent. For example, 3.0 x 10³ would be written, 3.0E3.
K=

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