Consider the following reaction occurring at 298K : N2O(g)+NO2(g)–>3NO(g)
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?
1 Answer
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I’m assuming it’s a mastering chemistry problem with a calculated standard free energy change. delta G’ of reaction.
For my variables I will use:
delta G’ as the standard free energy change, which is delta G knot of reaction
delta G as the free energy change, which is just delta G of reaction (no degree/knot symbol)
R= 8.314 J/mol*K
T= Temperature
Q=Kp= Partial Pressure rate law of the reaction
In this case, Kp= [P of NO]^3/[P NO2]*[P N2O]
If you have delta G’ of reaction (the standard change in free energy), and assuming that the reaction is not under standard conditions, you can use the equation delta G= delta G’ + R*T*ln (Kp)
The reaction is spontaneous when the free energy change is Negative (delta G of reaction). In order to have a negative free energy change, you need a negative value of R*T*ln (Kp) that is greater than your delta G’ (standard free energy change).
So therefore, -deltaG’= R*T* ln (Kp).
Say that you calculated a standard free energy change of 108,
then -108kJ= 8.314 (J/mol*K) *298K* ln([NO]^3/1).
Simplify, and you get -43.59= ln (x^3)
Raise “e” to the power of both sides: 1.17*10^-19= x^3
Simplify: x=4.89*10^-7 which is the partial pressure of NO in atm
I hope this helps:)
Source(s): Mastering Chemistry
