HClO(aq) + H2O (l) <> H3O+ (aq) + ClO-(aq)
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
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?
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
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.
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
Well, calculating ΔGrxn can be a bit tricky sometimes. It seems like you made a few mistakes in your calculations.
First, let's clarify the equation for Q. Q should be the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the balanced equation. In this case, the balanced equation is:
HClO(aq) + H2O(l) ↔ H3O+(aq) + ClO-(aq)
So, Q should be: [H3O+] * [ClO-] / [HClO]
Given that [ClO-] = [H3O+] = 5.69x10^-5 M and [HClO] = 1.340 M, we can calculate Q as:
Q = (5.69x10^-5)^2 / 1.340 ≈ 2.413x10^-9
Now, let's use the correct equation for ΔG:
ΔG = ΔGº + RTlnQ
Substituting the values, we get:
ΔG = 42.20 + (0.0083145)(298)ln(2.413x10^-9)
Calculating this expression, we get:
ΔG ≈ 42.20 + (2.472)(-20.57) ≈ 42.20 - 50.90 ≈ -8.70 kJ/mol
So, the correct value for ΔGrxn at 25 °C for hypochlorous acid is approximately -8.70 kJ/mol.
Remember, chemistry can be a bit tricky sometimes. But don't worry, it's all part of the fun! Keep up the good work!
It seems that there might be a small calculation error in your previous attempts. Let's go through the correct calculation step by step.
The equation you used is correct: ΔG = ΔGº + RTlnQ, where ΔGº is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25 + 273 = 298 K), and Q is the reaction quotient.
First, calculate lnQ. In this case, Q = [ClO-] / [HClO] = (5.69x10^-5) / (1.340) = 4.25x10^-5. Take the natural logarithm of Q:
lnQ = ln(4.25x10^-5) = -10.97
Now, calculate ΔG:
ΔG = ΔGº + RTlnQ = 42.20 kJ/mol + (8.314 J/(mol·K))(298 K)(-10.97)
= 42200 J/mol - 27817 J/mol
= 14383 J/mol
However, we need to convert the value to kJ/mol:
ΔG = 14383 J/mol / 1000 = 14.38 kJ/mol
So, the correct value of ΔG for the reaction of hypochlorous acid at 25 °C is 14.38 kJ/mol.
It's essential to double-check your calculations to avoid small mistakes like sign errors or incorrect logarithm calculations.