Thermo homework answers by 50A1Mpa2

VIEWS: 6 PAGES: 4

									1981 D

Four effects and 4 explanations valued at 1 point each.
a) No, the reactant is favored. Treaction is endothermic and products are at highter energy
than the reactant.

b) Yes, products are favored. [delta]S > O as 2 moles of gas are produced form 1 mole of
gas.

c) Right, products are favored. For the system at constant pressure: absorption of heat
favors the products. An argument using LeChatelier's principle can be used with heat
considered as a reactant; or T[delta]S favors products as [delta]G becomes more negative.
For the system at constant volume: products are favored but increase in the yield of
products results in an increase in pressure which would drive the reaction to the left.
No points were deducted for students who mentioned both the thermodynamic effect and
the counteractiong pressure effect at constant volume but were uncertain as to which
effect is larger. Students who considered a system at constant volume and stated only that
the increase in pressure would drive the reaction to the left received half credit.

d) Left, the reactant is favored. A decrease in the volume would increase the pressure and
the strain is relieved by the reverse reaction which produces 1 mole of gas from 2 moles
of gas.



1983 B


2)
a) two points
[delta]G° = [sigma] [delta]Gf° (prod) - [sigma] [delta]Gf° (react)
[delta]G° = - 166.2 - (- 137.3 + 2 (0))
[delta]G° = -28.9 KJ/mol

b) two points
[delta]G° = - RT ln K (or - 2.3 RT log K)
- 28.9 = - ( 8.31 x 10¯3) (298) ln K
ln K = 11.67
K = 1.17 x 105

c) two points
[delta]G° = [delta]H° - T [delta]S°
- 28,900 = -128,100 - 298 [delta]S°
[delta]S° = - 99,200 / 298 = - 333 J / mol-K

d) three points
[delta]Hf° (H2) = 0 and [delta]Gf° (H2) = 0
deltaS° = [sigma]S° (prod) - [sigma]S° (react)
- 333 J / mol-K = 126.8 J / mol-K - 197.9 J / mol-K - 2 S° (H2)
S° = 131 J / mol-K

1984 B
Average score 5.52
a) three points
From Hess's law:
[delta]H°f = [4 (393.5) + 4 (205.85)] - 2183.5 kJ = - 533.8 kJ


b) one point
4 C(s) + 4 H2(g) + O2(g) ---> C3H7COOH


c) two points
[delta]S°f butyric acid = S° butyric acid - 4 S° carbon - 4 S° H2 - S° O2
[delta]S°f butyric acid = [226.3 - 4(5.69) - 4(130.6) - 205] joules/K = - 523.9 joules/K
Note: If part c was based on the equation in part b, and was done correctly, credit was
given for part c even if part b was wrong.


d) three points
[delta]G°f butyric acid = [delta]H°f - T [delta]S°f
[delta]G°f = [- 533.9 - (298) (- 0.524)] kJ
[delta]G°f= - 377.7 kJ
Note: For each of the problems, a maximum of one point was subtracted for gross misuse
of significant figures and. or for a mathematical error if correct principles were used.


1992 B
3) average = 3.5 (Only 30 scores of nine; kids did not see stoichiometry in (b), had
problems on which gas constant to use, and a hard time in (c) in relating a more negative
value.)
a) two points [delta]G° = - RT ln K; rearranging gives ln K = [delta]G° ÷ - RT ln K = -
123,000 J ÷ - ((8.31 J/mol K) (298 K)) = 49.7 K = 3.72 x 1021


b) two points
[delta]G° = [delta]H° - T[delta]S°
- 246,000 J = - 326,400 J - (298) (x)
x = - 270 J / K


c) two points
[delta]S is a larger negative number
ClF3 (liquid) is more ordered (less disordered) than ClF3 (gas)
This was my answer before I saw the standard given above: A liquid is more ordered than
a gas. There would be a greater entropy change in gas + gas ---> liquid than in gas + gas -
--> gas. Therefore, sign is the same, but absolute magnitude is greater.


d) three points

i) ClF3 is a more complex molecule (i.e. more atoms) with more vibrational and rotational
degrees of freedom than Cl2

ii) Cl2 + 3 F2 ---> 2 ClF3; use Hess's Law
[sigma]Srxn = [sigma]S products - [sigma]S reactants
- 270 = [2 (281.50)] - [222.96 + 3x]
x = 203 J mol¯1 K¯1


1997D


(a) S° is positive (or "+", or ">0") 1 point
Moles products > moles reactants 1 point
Note; all species are gaseous, so (g) need not be indicated. To earn credit, number of
particles (moles) must be discussed. No explanation point earned for just nothing that
disorder increases, or that PCl5 is decomposing or dissociating.
(b) G° will decrease (or become more negative, or become smaller). 1 point
G° = H° - TS°
and since S° is positive, TS° is positive ( > 0). Thus increasing T will result in a larger
term being subtracted from H°, or, G° = -RT ln K and K is going up in value since T is
increasing.)
Note: full credit earned for part (b) if:
S° < 0 in part (a) which leads to G° is increasing because TS° is added to H°, or,
S° = 0 in part (a) which leads to no change in G°
(c) no change (one point)
PHe is not part of the a) reaction (He is not involved) or, b) law of mass action or, c)
reaction quotient or, d) equilibrium constant expression; one point
hence altering PHe has no effect on the position at equilibrium
(d) moles of Cl2 will decrease (one point)
The decrease in volume leads to an increase in pressure (concentration), therefore the
reaction shifts to the left because:
(one point for any of the following)
Q > Ksp (Q > Kc, or,
the rate of the reverse reaction increase more than the rate of the forward reaction, or,
the reaction shifts toward the lesser moles of gas.
Note: "LeChatelier's principle" alone is not sufficient to earn the explanation point. If
response suggests that the number of moles of Cl2 is halved because the system is "cut" in
half, only one point is earned.

								
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