I'm studying for my final in Chem 2 and I am studying one of the sheets that the professor gave us. It has all the answers so I'm not looking for the answers, but more of the thought process of how you get to the answer. You can do all of them, one of them, or half of one. Any help is greatly appreciated!

for 21. look at your equations and see what you can calculate. you can do all apart from speed (requires kinetics)

equilibrium constant from equation in q24.
extent of rxn - related to equilibrium
direction of rxn - whether delta G is negative for forward or reverse rxn
temperature of spontaneous rxn - can be calculated using equation from q23.

for 22. just do sum of products - sum of reactants

(-1353 + -394.4) - (-1139 + - 370) = -238.4 kjmol-1

for 23. do the same as 22 for both delta H and delta S (sum of products - sum of reactants) then substitute them into the equation given, converting degrees to kelvin by +273.15

for 24. just put delta G into the equation

for 25. higher emf = stronger oxidising agent, lower emf = stronger reducing agent

for 26. strange question - it seems like u would balance it similarly to a redox equation. Cr2O7- has the Cr in +6 oxidation state going to the +3 oxidation state (twice) so u need +6 charge to balance - which is done with 3Ni2+. its 2Cr3+ because Cr2. what a pathetic explanation to an awkward question - if you need more help pm me ill try explain better

for 27. part 1 is just the reverse so its positive instead of negative. part 2 is higher stoichiometry but it doesnt affect the emf - can simplify stoichiometry to original equation given so its same -> -0.76

if you dont understand anything just pm me and ill try give better explanations

adding on a little bit here

27)

Like the name suggests, EMF values are always given for chemical reductions that are REACTIONS. Remember that

Leo the lion says GER

LEO GER

Losing Electrons Oxidation
Gaining Electrons Reduction

So you're given a reduction potential from a table, what do you do if they ask you for the oxidizing value? Like Arnold said, reverse it. Ecell(oxidation) = +0.76


The second part is a trick question. Stoichiometry / higher coefficients don't affect the EMF.

I did 23 and got like 37000 for the answer using that method. Are you sure that is correct? Also do I need to use the coefficients?

Worked through the first two and got those right so far.

Also I don't really understand this stuff so a little more detail/explaining would be extremely helpful.

Just did 24 and got 1.0067 instead of the 8.12 x 10^2.

This post was edited by Braxton11 on Dec 11 2013 02:16pm

Yes, use the coefficients; it's the only way it'd make sense
reaction specific

I used the coefficients and got delta H = 31.1 and delta S = 66.3

Delta G = 31.1 - (298K)(66.3) = 1657.5

The number is still way too high. What am I doing wrong.

gibbs dalton law apply

It's the sum up your products - reactants

so it'll look like
Delta G =( 2H_ag +.5H_O2 - H_ag2o ) - T (2S_ag +.5S_O2 - S_ag2o)

This post was edited by saber_x3 on Dec 11 2013 02:52pm

For some reason this makes absolutely no sense to me. Would you be able to word this differently please.

this is basically what i said

products - reactants (include stoichiometric coefficients) so lets calc delta H first.

Products = 2*0 + 1/2*0 = 0 (zero by definition for elements)

Reactants = -31.1 KJmol-1

0 - -31.1 = 31.1 KJmol-1

now lets calc delta S.

Products = 2*42.55 + 1/2*205 = 187.6 J mol-1 K-1

Reactants = 121.3 J mol-1 K-1

187.6 - 121.3 = 66.3 J mol-1 K-1

lets calc T*delta S = 298.15 * 66.3 = 19767.345 J mol-1

divide by 1000 to get KJmol-1 = 19.767 KJ mol-1 <---- you missed this part

31.1 - 19.767 = 11.333 KJ

This post was edited by ArnoldChlamydia on Dec 11 2013 06:35pm

for 24, convert KJmol-1 to Jmol-1 by x1000

-16600 Jmol-1

Kp = e^(- (-16600) / 8.314 * 298)

= 812 = 8.12*10^2