Asked by Tyler
The following reaction has an equilibrium constant (Keq) of 160.
2 NO2(g) <--> 2 NO(g) + O2(g)
What will be the reaction quotient (Q) and in which direction will the reaction proceed if the partial pressure of NO2 is 5.0*10^-4 atm, NO is 0.080 atm, and O2 is 0.020 atm?
Work:
Keq= [NO]^2 [O2]/[NO2]
Keq=[0.080]^2 [.020]/ 5.0*10^-4
Keq=1.28*10^-4/5.0*10^-4
Keq=.256 which is less than 1 meaning that the reactants are favored moving the reaction towards the left.
I am not sure if .256 is also the reaction quotient or if it's a totally different process to find it.
2 NO2(g) <--> 2 NO(g) + O2(g)
What will be the reaction quotient (Q) and in which direction will the reaction proceed if the partial pressure of NO2 is 5.0*10^-4 atm, NO is 0.080 atm, and O2 is 0.020 atm?
Work:
Keq= [NO]^2 [O2]/[NO2]
Keq=[0.080]^2 [.020]/ 5.0*10^-4
Keq=1.28*10^-4/5.0*10^-4
Keq=.256 which is less than 1 meaning that the reactants are favored moving the reaction towards the left.
I am not sure if .256 is also the reaction quotient or if it's a totally different process to find it.
Answers
Answered by
Tyler
Nevermind I figured it out
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