Calculate the G and the equilibrium constant for the following reaction at 320K.

ID: 881392 • Letter: C

Question

2 NO(g) + O2(g) <--> 2NO2(g)

Hf (kJ/mol) Gf (kJ/mol) S (J/mol·K)

NO(g) 91.3 87.6 210.8

O2(g) 0 0 205.2

NO2(g) 33.2 51.3 240.2

1/2 N2(g) + O2(g) --> NO2(g), deltaH = 33.2 kJ

we need to multiply the above equation by 2 in order to get the desired coefficients

N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ

1/2 N2(g) + 1/2O2(g) --> NO(g), delta H = 91.3 kJ

we need to multiply the above equation by 2 and invert it in order to get the desired coefficients

2 NO -> N2 + O2..............deltaH = -182.6 kJ

Delta H = 66.4 - 182.6 = -116.2 KJ/mol

Similarly we can calculate the entropy of the reaction

Entropy of reaction = 2 * (entropy of NO2) - (entropy of O2 + 2*entropy of NO)

= -146.4 J/mol-K

Delta G (320K) = Delta H - T(Delta S)

= -116.2 KJ/mol + (146.4)(320)

= -69.352 KJ/mol

Now for finding equilibrium constant

-69.352 * 1000 = -(8.314)(320)ln(Keq)

ln(Keq) = 26.0674

Keq = e^(26.0674) = 2.31 * 10^(11)

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