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
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.
Is this right?
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
a) To write the equilibrium expression, we need to denote the concentrations of the reactants and products in their balanced stoichiometric coefficients. The equilibrium expression is written in the form of:
K = [N2O4]^a/[NO2]^b
Where:
- K is the equilibrium constant.
- [N2O4] is the concentration of N2O4.
- [NO2] is the concentration of NO2.
- a and b are the coefficients of the species N2O4 and NO2, respectively, in the balanced chemical equation.
In this case, the balanced equation is:
2NO2(g) <--> N2O4
Therefore, the equilibrium expression is:
K = [N2O4]/[NO2]^2 (since a = 1 and b = 2 in this case)
b) To calculate the concentration of N2O4(g) present in equilibrium with 0.50 mole of NO2, we assume that the reaction has reached equilibrium. Since we know the equilibrium constant (K) and the initial concentration of NO2, we can use the equilibrium expression to find the concentration of N2O4.
Given:
Initial concentration of NO2, [NO2] = 0.50 moles
Let's denote the equilibrium concentration of N2O4 as [N2O4]eq.
Using the equilibrium expression:
K = [N2O4]eq/[NO2]^2
Now we can rearrange the equation to solve for [N2O4]eq:
[N2O4]eq = K * [NO2]^2
Substituting the values:
[N2O4]eq = 1.15 * (0.50)^2 = 1.15 * 0.25 = 0.2875 moles
Therefore, the concentration of N2O4(g) in equilibrium with 0.50 mole of NO2 is 0.2875 moles.