the extra NO being added threw me off..
or is it supposed to be like Kc = 0.800M^2/[0.100M][0.100M] = 64 M?
At equilibrium, the concentrations in this system were found to be [N2] = [O2] = 0.100 M and [NO] = 0.500 M.
N2(g) + O2(g) <-> 2NO(g)
If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?
Kc = 0.800 M/[0.100M][0.100M] = 80 M?
7 answers
No and no.
First, since the values quoted are equilibrium values, we need to calculate Kc (by the way, 80 is not an equilibrium value).
............N2 + O2 ==> 2NO
equil.......0.1..0.1.....0.5
Kc = (NO)^2/(N2)(O2)
Kc = (0.5)^2/(0.1)(0.1)
Kc = 25 is what I have.
Then we set up another ICE chart.
............N2 + O2 ==> 2NO
initial.....0.1...0.1....0.8
change.....+x.....+x.....-2x
equil......0.1+x..0.1+x...0.8-x
Kc = 25 = (NO)^2/(N2)(O2)
25 = ((0.8-2x)^2/(0.1+x)(0.1x)
Solve for x.
By the way, you need NOT go through a quadratic if you notice that you can take the square root of both sides.
First, since the values quoted are equilibrium values, we need to calculate Kc (by the way, 80 is not an equilibrium value).
............N2 + O2 ==> 2NO
equil.......0.1..0.1.....0.5
Kc = (NO)^2/(N2)(O2)
Kc = (0.5)^2/(0.1)(0.1)
Kc = 25 is what I have.
Then we set up another ICE chart.
............N2 + O2 ==> 2NO
initial.....0.1...0.1....0.8
change.....+x.....+x.....-2x
equil......0.1+x..0.1+x...0.8-x
Kc = 25 = (NO)^2/(N2)(O2)
25 = ((0.8-2x)^2/(0.1+x)(0.1x)
Solve for x.
By the way, you need NOT go through a quadratic if you notice that you can take the square root of both sides.
why does the equil end up being 0.8-x instead of 0.8 - 2x? and i solved for x but it gave me 2 different answers did I do something wrong?
First, it wasn't -2x because I made a typo.It's obvious from the initial value and the change value that it SHOULD be 0.800-2x and, in fact, I wrote (0.800-2x)^2 when I substituted into the K expression.
You ALWAYS get two answers when you solve a quadratic. Most of the time it is obvious which is the right one and which is the wrong one. For example, I would expect, although I didn't go through the math, if you take the two x values you have, multiply by 2 and subtract from 0.800 that one will be a positive number and one a negative number. The one giving a negative number when subtracted from 0.800 will be the one to throw away since you can't have negative concns.
You ALWAYS get two answers when you solve a quadratic. Most of the time it is obvious which is the right one and which is the wrong one. For example, I would expect, although I didn't go through the math, if you take the two x values you have, multiply by 2 and subtract from 0.800 that one will be a positive number and one a negative number. The one giving a negative number when subtracted from 0.800 will be the one to throw away since you can't have negative concns.
Cngkreng jkt .
I don't know how to do the last algebra part that DrBob22 did not explain.
I have not taken algebra since 2 years ago it would have been helpful to see...
I haven't found out the math part anywhere!
I have not taken algebra since 2 years ago it would have been helpful to see...
I haven't found out the math part anywhere!
How do we construct an ICE chart for the concns of the dissociated ions to form a precipitate so we can calculate the final concns of ions in solution?
2 answers