i already posted this question, but im still confused on it.
Dinitrogentetraoxide partially decomposes according to the following equilibrium:
N2O4 (g) -> 2NO2 (g)
<-
A 1.00-L flask is charged with 0.400 mol of N2O4. At equilibrium at 373 K, 0.0055 mol ofN2O4
remains. Keq for this reaction is ?.
so my steps:
N2O4 (g) -> 2NO2 (g)
Initial conc. 0.400 mol 0 mole (no reaction yet)
Change -0.3945 mol + 2 x 0.3945 mole (based on equation above)
Equilibrium 0.400-0.3945 mol 0.7890 mole
0.0055 mol 0.7890
Keq = [NO2]2/[N2O4]
Keq = (0.789)^2/(0.0055) =113
but my book says the answer is 0.87?
I think the book answer must be wrong. I used 113.2 as Keq and started with 0.4 M N2O4 and zero NO2 and solved backwards. x came out to be 0.3945M which makes equilibrium concn N2O4 = 0.0055 which was our starting number in the problem.
If I use 0.87 as Keq, the quadratic has a solution which working backwards gives x = 0.2056 so (N2O4) at equilibrium = 0.1944 (not 0.0055). The values of 0.2056 for (N2O4) and 0.411 = (NO2) gives Keq = 0.87; however, the problem gives (N2O4) at equilibrium of 0.0055 M and not 0.2056M.
I don't believe 0.87 is correct.
yes, ive asked multiple people and they've all told me the same thing. i'll study the question the way you said, i have a test next week and i wanted to know how to solve problems like this,
To solve this problem, let's calculate the equilibrium concentrations of N2O4 and NO2 using the given information:
Initial concentration of N2O4 = 0.400 mol
Change in concentration of N2O4 = 0.400 mol - 0.0055 mol = 0.3945 mol
Equilibrium concentration of N2O4 = 0.0055 mol
According to the balanced equation, for every 1 mole of N2O4 that reacts, 2 moles of NO2 are produced. Therefore:
Change in concentration of NO2 = 2 * change in concentration of N2O4 = 2 * 0.3945 mol = 0.7890 mol
Equilibrium concentration of NO2 = 0.7890 mol
Now, let's calculate the equilibrium constant Keq using the equilibrium concentrations:
Keq = [NO2]^2 / [N2O4]
Keq = (0.7890 mol)^2 / (0.0055 mol)
Keq ≈ 113.52
So, based on the calculations, the equilibrium constant Keq for this reaction is approximately 113.52, not 0.87 as mentioned in the book.
The steps you have followed are correct, but it seems like there was a small calculation mistake in your calculation of the equilibrium concentrations.
Let's re-calculate the equilibrium concentrations:
Using the chemical equation:
N2O4 (g) --> 2NO2 (g)
Initial concentration of N2O4 = 0.400 mol
No initial concentration of NO2 since the reaction hasn't started yet.
Change in concentration:
Since 0.0055 mol of N2O4 remains at equilibrium, we can calculate the change in concentration of N2O4:
Change in N2O4 = Initial concentration - Equilibrium concentration = 0.400 - 0.0055 = 0.3945 mol
According to the stoichiometry of the equation, the change in concentration of NO2 will be twice the change in N2O4:
Change in NO2 = 2 × 0.3945 = 0.7890 mol
Equilibrium concentrations:
Equilibrium concentration of N2O4 = Initial concentration - Change in N2O4 = 0.400 - 0.3945 = 0.0055 mol
Equilibrium concentration of NO2 = Change in NO2 = 0.7890 mol
Now let's calculate the Keq using these equilibrium concentrations:
Keq = [NO2]^2 /[N2O4]
= (0.7890)^2 /(0.0055)
= 112.874
Therefore, the correct value of Keq for this reaction is approximately 112.874, not 0.87 as mentioned in your book.