Homework Statement
For the reaction
2 NO_{2}(g) \leftrightarrow N_{2}O_{4}(g)
ΔH°_{rxn} = -57.2 kJ and ΔS°_{rxn} = -175.8 J/K
A 1 L container is initially filled with 1 atm of NO_{2} at 298 K. Find the equilibrium pressures of NO_{2} and N_{2}O_{4} at 298 K if
a. volume is held...
Yep! This was actually a very DOH! moment for me.
So I forgot that I can just plug in the equation of state and write P in terms of V.
This leads to the differential equation:
-\intdA = \int\frac{nRT}{V}-\frac{an^{2}}{v^{2}}dV,
from 20L to 40L
This comes out as
nRT...
Homework Statement
The 4 fundamental equations of thermodynamics are:
dE = TdS - PdV
dH = TdS + VdP
dG = VdP - SdT
dA = - PdV - SdT
Supose a gas obeys the equation of state
P = \frac{nRT}{V} - \frac{an^{2}}{V^{2}}
Use one of the fundamental equations to find the change in Helmholtz energy...
Ah, yes. I apologize for that. I'll be more specific next time.
This is a made up problem that goes along with a lab assignment.
To elaborate, the 0.2 N HCl information goes along with my titration curve below:
As far as I know, the actual amino acid unknown solution was prepared...
Sorry, it seems the 0.2 N HCl information only applies to my actual lab data. The question above is what I assume to be a practice calculation to be done with my lab data
Alright well, I've tried something else. Starting with the equation for the reaction:
H2X + 2NaOH → Na2X + 2H2O
Where X is the amino acid. The equation states that 2 moles of base titrates 1 mole of amino acid.
Assuming the change from pH 0.8 to pH 12.0 is the full titration (meaning the both...
I probably should have specified that. The amino acid was dissolved in 0.2 N HCl. So the amine group is probably protonated, and the overall charge is +1
No. It should have at least two titratable groups
Homework Statement
7.39 mmol of base was used when titrating 432 mg of a monoamine monocarboxylic amino acid from pH 0.8 to 12.0. What is the name of the amino acid?
Homework Equations
None I suppose. Henderson-Hasselbalch perhaps, but I don't see how that helps me
The Attempt at...
Ah yes, thank you.
I have one more item I'd like to clarify.
I have two equation for thermodynamic work listed in my notes.
1. w = -∫PdV
2. w = -nRT ln\frac{V_{2}}{V_{2}}
I used the first equation to obtain -1013J, but the second equation provides a different value of -987.7J. This...
Okay, so what I'm getting is that since it's an isothermal process, the change in internal energy is 0.
That would mean q is just the reverse of w. Does this sound right? That would mean this is reversible too, yes?
Homework Statement
One mole of ideal gas expands isothermally at 20°C against a constant pressure of 1 atm from 20 to 30L. (1 atm = 1.013 x 105Pa; 1 m3 = 1000L
Find q, w, ΔE, and ΔH in Joules
Homework Equations
ΔE = q + w
w = -\intPdV
H = E + PV
The Attempt at a Solution
I solved for work...
Ah, I see now. It's been a while since I've done that, but it works. I suppose I should do similar exercises to get this in my head.
And yeah, part of the integral turned out to be an inverse tangent one. Thanks a lot.
Homework Statement
\int\frac{8x^{2}+5x+8}{x^{3}-1}
Homework Equations
Because the denominator can be reduced to (x-1)(x^{2}+x+1), I set up the partial fractions to be \frac{A}{(x-1)} + \frac{Bx+C}{(x^{2}+x+1)}
The Attempt at a Solution
I've solved for A, B, and C, and now have...