HClO(aq) + H2O (l) <> H3O+ (aq) + ClO-(aq)

Question

Calculate the value of ΔGrxn at 25 °C for hypochlorous acid when

[ClO-]=[H3O+]=5.69x10^-5 M
[HClO] = 1.340 M

Previously calculated a ΔGºrxn of 42.20 kJ/mol

Q = [5.69x10^-5] / [1.340] = .00004245 M

equation used:
ΔG = ΔGº + RTlnQ
ΔG = 42.40 + (.008315)(298)ln.00004245
ΔG = 42.40 + (2.468)(-10.07)
ΔG = 42.40 + (-24.95)
ΔG = 42.40 - 24.95
ΔG = 17.45

It was incorrect. :(

I also tried
ΔG = RTlnQ
ΔG = (.0083145)(298)ln.00004245
ΔG = -24.94
Incorrect.

Thanks for any and all help <3

Answer 1

Two comments.
Shouldn't that Q be (5.69E-5)^2/(1.34)?
Can you add dGo (in kJ) and RTlnQ (which I believe is in J) without convert one or the other?

Answer 2

I usually convert it (especially in ΔG = ΔGº + RTlnQ ) because it will confuse me other wise and I will get wrong numbers. I usually keep kJ because that's usually what I need, unless specified otherwise.

I suppose I don't need to convert it for ΔG = RTlnQ.

And why would it 5.69X10^-5 be to the second power? Is the chemical equation imbalanced? or is it because theres two compounds that equal said number?

[ClO-]=[H3O+]=5.69x10^-5 M

Answer 3

For the reaction A + B ==> C + D
Q = (C)(D)/(A)(B) which looks like Keq but isn't. I suppose the simple answer is yes. You have two components of equal magnitude on the right and those are the products.

Answer 4

okay so then

Q = [5.69x10^-9]^2/1.3405 =
3.238x10^-9/1.3405 = 2.42x10^-9

ΔG = 42.20 + (.00833145)(298) ln (2.42x10^-9)
= 42.40 + (2.477) (-19.84)
= 42.40 + -49.144
= -6.74 rounded to 7 which is correct.

Thank you sir! <3