Could you please explain the solution to #28. The answer should be 3.02*10^12. 2

Question

Could you please explain the solution to #28. The answer should be 3.02*10^12.

27) Consider the following reaction at 298 K: @s-71.2 kJ 2 NO (g) + 02 (g) 2 NO2 (g) Calculate G for the following non-standard conditions: 7: 450 K P NO 0.100 atm PO2 = 0.120 atm P NO2 = 1.75 atm Is the reaction more spontaneous or less spontaneous at these non-standard conditions compared to standard conditions? A) 41.8 kJ, less spontaneous Ens 98.3 kJ, more spontaneous 41.8 kJ, more spontaneous +41.8 k], less spontaneous E) -98.3 kJ, less spontaneous Q- 9552.08 28 Forthe reaction in question #27, calculate the equilibrium constant at 298 K 2 A)4.44 x 1011 B) 9.31 x 109 C) 3.31 x 10-13 D) 3.02 x 1012 E) 6.23 x 10-10 Da-YL3KS 29) Gibb's Free Energy (A G), for a chemical reaction is defined as: A) The amount of energy required to overcome negative entropy changes in the system. B The amount of energy required to make the reaction proceed.

Explanation / Answer

28)

T = 298 K

deltaG = -71.2 KJ/mol

deltaG = -71200 J/mol

we have below equation to be used:

deltaG = -R*T*ln Kc

-71200 = - 8.314*298.0* ln(K)

ln K = 28.7378

K = 3.02*10^12

Answer: 3.02*10^12

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