Don't you know Keq expression? products/reactants with coefficients as exponents.
b. Substitute 0.40 mol for NO2 into Keq and solve for N2O4.
At 55C, the K for the reaction: 2NO2(g) <--> N2O4 is 1.15
a) write the equilibrium expression
b) calculate the concentration of N2O4(g) present in equilibrium with 0.50 mole of NO2
Please help and explain
5 answers
Is this right?
a) kc= [N2O4]/[NO2]^2
=(1.15)/(2)^2
=0.2875?
B) I still can't figure it out :( please help
a) kc= [N2O4]/[NO2]^2
=(1.15)/(2)^2
=0.2875?
B) I still can't figure it out :( please help
Frankly I don't know how to answer the question. Is this Kc or is it Kp? The problem calls it Keq and gives moles. Moles can't go in the Keq expression. It must be molarity for Kc or partial pressure for Kp. With no volume given mols can't be converted to M nor to pressure. Is there more to the problem or did you post an edited version? The Kc expression you wrote is perfect.
Okay thank you
a) Kc= [N2O4]/[N02]^2
b) first find the concentration for NO2
write down what is given
n= 0.50mol
v=1L
c=?
c=n/v sub in the values 0.50mol/1L= 0.50mol/L
Lastly we sub this concentration to the kc expression in a.
the kc is given which is 1.15
1.15= [N2O4]/[0.50]^2
1.15= [N204]/[0.25]
then we times the denominator which is 0.25 by 1.15 and we get 0.2875mol/L. There ya go princess
b) first find the concentration for NO2
write down what is given
n= 0.50mol
v=1L
c=?
c=n/v sub in the values 0.50mol/1L= 0.50mol/L
Lastly we sub this concentration to the kc expression in a.
the kc is given which is 1.15
1.15= [N2O4]/[0.50]^2
1.15= [N204]/[0.25]
then we times the denominator which is 0.25 by 1.15 and we get 0.2875mol/L. There ya go princess
3 answers