Question

Use the table below to answer question for the following reaction.

N2O(g)+12O2(g)⇌2NO(g)

Species	ΔfG∘,kJ/mol
NO(g)	86.55
N2O(g)	104.2
O2(g)	0.00 A)Establish ΔrG∘ at 298 K

B)Establish K at 298 K for the reaction

Answer 1

A)

we have:

Gof(N2O(g)) = 104.2 KJ/mol

Gof(O2(g)) = 0.0 KJ/mol

Gof(NO(g)) = 86.55 KJ/mol

we have the Balanced chemical equation as:

N2O(g) + 1/2 O2(g) ---> 2 NO(g)

deltaGo rxn = 2*Gof(NO(g)) - 1*Gof( N2O(g)) - 1/2 *Gof(O2(g))

deltaGo rxn = 2*(86.55) - 1*(104.2) - 1/2 *(0.0)

deltaGo rxn = 68.9 KJ/mol

Answer: 68.9 KJ/mol

B)

T = 298 K

G = 68.9 KJ/mol

G = 68900 J/mol

we have below equation to be used

deltaG = -R*T*ln K

68900 = - 8.314*298.0* ln(K)

ln K = -27.8095

K = e^(-27.8095)

K = 8.37*10^-13

Answer: 8.37*10^-13