Consider the following reaction occurring at 298 K: N2O(g)+NO2(g)3NO(g) Part A S

Question

Consider the following reaction occurring at 298 K: N2O(g)+NO2(g)3NO(g)

Part A Show that the reaction is not spontaneous under standard conditions by calculating Grxn.

Part B If a reaction mixture contains only N2O and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until some NO forms in the mixture. What maximum partial pressure of NO builds up before the reaction ceases to be spontaneous?

Part C Can the reaction be made more spontaneous by an increase or decrease in temperature? Yes or No

Explanation / Answer

Q1.

N2O(g)+NO2(g)3NO(g)

dG = Gprodcts- Greactants

dG = 3*86.56696 - (104.1816 + 51.29584) = 104.22344 kJ/mol

since dG > 0, this must not favour producs, i.e. it is non.spontaneous

Q2.

dG = -RT*ln(Kp)

Kp = exp(-dG/(RT))

Kp = exp(-104223.44/(8.314*298)) = 5.37818739*10^-19

Kp = NO^3 / (N2O)(NO2)

initially

NO= 0

N2O = 1

NO2 = 1

in equilibrium

NO= 0 + 3x

N2O = 1 - x

NO2 = 1 - x

substitute

5.378 *10^-19 = (3x)^3 / (1-x)(1-x)

(5.378 *10^-19)(1 -2x +x^2) = 27x^3

x = partially 0 atm

therefore

Almost no NO3 will be zero

Part C Can the reaction be made more spontaneous by an increase or decrease in temperature? Yes or No

Not possible, since

dG = dG º + RT*ln(Q)

therefore, even though we can decrease T, it will never be negative, meaning tha this will be never spontaneous

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