Reviewing Thermo

Chem I | Chem II AP
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1)
Which of the following in each pair has the higher positional entropy?

Positional entropy is the same as entropy of the system or ΔSsys.

a)

A gas has more entropy than an aqueous solution.

b)

A liquid has more entropy than a solid.

c)

A gas at a decreased pressure has more entropy because a decreased pressure is caused by increased volume. If the volume of the container is greater there are more places in the container for the molecules to fill.

d)

N2O4 has greater molecular complexity than NO2. There are more ways to arrange 6 atoms than 3 atoms.

2)
Predict the sign of the entropy change for each of the following:
a)
KOH pellets dissolve in water.

Solid → aqueous solution

b)
Solid ammonium dichromate burns to produce water vapor and nitrogen gas.

Solid → gas

c)
Saturated solution of calcium acetate is mixed with ethanol to form a gel.

Aqueous solution → solid

3)
Calculate ΔSsurr for each reaction at 25°C and 1 atm.
ΔSsurr =
- ΔH
T
a)
Br(l) → Br(g)
 
ΔH = 31 kJ
ΔS =
J/K
ΔSsurr =
- ΔH
T
=
- (31 kJ)
298 K
= - 0.104 kJ/K x
1000 J
1 kJ
= -104 J/K
b)
Combustion of C2H6
 
ΔH = - 2857 kJ
ΔS =
J/K
ΔSsurr =
- ΔH
T
=
- (- 2857 kJ)
298 K
= 9.587 kJ/K x
1000 J
1 kJ
= 9587 J/K
4)
A chemical reaction gives a change in entropy of the universe of -48 J/K. Is the process spontaneous?
Why or why not?
5)
Which of the following represent spontaneous processes? Which are exothermic?

A reaction is spontaneous if ΔSuniv is increasing. A reaction is exothermic if ΔSsurr is + because of the formula -ΔH/T.

a)
ΔSsys = 358 J/K
 
ΔSsurr = 420 J/K
 

Spontaneous because ΔSuniv = 778 J/K (a positive number); Exothermic because ΔSsurr = 420 J/K (a positive number)

b)
ΔSsys = - 358 J/K
 
ΔSsurr = - 52 J/K
 

Nonspontaneous because ΔSuniv = -410 J/K (a negative number); Endothermic because ΔSsurr = - 52 J/K (a negative number)

c)
ΔSsys = - 358 J/K
 
ΔSsurr = 463 J/K
 

Spontaneous because ΔSuniv = 78 J/K (a positive number); Exothermic because ΔSsurr = 463 J/K (a positive number)

6)
Determine whether each of the following sets of data represent spontaneous or nonspontaneous proesses.

You must calculate ΔG if ΔH & ΔS have the same sign to see if the temperature is high enough (or low enough) for the reaction to be spontaneous.

a)
ΔH = - 16 kJ
 
ΔS = 50 J/K
 
T = 300 K
 

This reaction will always be spontaneous because ΔH is - and ΔS is +.

b)
ΔH = - 5 kJ
 
ΔS = - 20 J/K
 
T = 200 K
 
ΔG = ΔH - TΔS = -5 kJ - (200K)(-.02kJ) = -1 kJ
Spontaneous because ΔG is negative.
c)
ΔH = - 5 kJ
 
ΔS = - 20 J/K
 
T = 500 K
 
ΔG = ΔH - TΔS = -5 kJ - (500K)(-.02kJ) = 5 kJ
Nonspontaneous because ΔG is positive.
d)
ΔH = 10 kJ
 
ΔS = - 20 J/K
 
T = 200 K
 

This reaction will never be spontaneous because ΔH is + and ΔS is -.

7)
Predict the sign of the entropy change for each of the following.
a)
evaporating ethanol at room temperature

liquid → gas

b)
cooling nitrogen from 80°C to 20°C

temperature decrease

c)
freezing bromine

liquid → solid

8)
The heat of fusion for element X is 10.50 kJ/mol. The entropy of fusion is 9.6 J/Kmol. Calculate the melting point of element X.
°C
At a phase change, the reaction is at equilibrium so ΔG = 0 = ΔH - TΔS
Rearranging that formula will give:T =
ΔH
ΔS
=
10.50 kJ
0.0096 kJ
= 1093.8 K = 820.8°C
9)
The normal boiling point of diethyl ether is 308 K. The enthalpy of vaporization is 27.2 kJ/mol. Calculate ΔS for the vaporization of 1.0 mol of diethyl ether.
J/K
At a phase change, the reaction is at equilibrium so ΔG = 0 = ΔH - TΔS
Rearranging that formula will give:ΔS =
ΔH
T
=
27200 J
308 K
- 88.3 J/K
10)
Given the following data:
2 H2(g) + C(s) → CH4(g)
 
ΔG = - 51 kJ
2 H2(g) + O2(g) → 2 H2O(l)
 
ΔG = - 474 kJ
C(s) + O2(g) → CO2(g)
 
ΔG = - 394 kJ
 
Calculate ΔG for:
CH4 + 2 O2 → CO2 + 2 H2O
ΔG =
kJ

Rearrange the equations to cancel out unwanted species. Change the ΔG values accordingly. Add all the equations and ΔG values.

CH42 H2+C  ΔG = 51 kJ
2 H2+O22 H2O  ΔG = -474 kJ
C+O2CO2 +ΔG = -394 kJ
CH4 + 2 O2 → CO2 + 2 H2O ΔG = -817 kJ
11)
If R = 1.99 cal/molK, calculate K for the dissociation of HCl at 25°C given that:
 
1/2 H2 + 1/2 Cl2 → HCl
 
ΔG = -22.7 kcal
K =
ΔG = - RT ln(K)
-22.7 kcal = - (0.00199 kcal)(298 K) ln(K)
-38.3 = ln(K)  K = 2.4 x 10-17
What does this say about the tendency of HCl to dissociate?

Because the K value is less than 1, the reaction is nonspontaneous.

12)
The value of the equilibrium constant for a given reaction is K = 6 x 10-23.
What does that indicate about the spontaneity of the reaction?

Because the K value is less than 1, the reaction is nonspontaneous.

13)
The value of the equilibrium constant for a given reaction is K = 8 x 1058.
What does this tell us regarding the speed of the reaction?

The K value will indicate if the reaction will happen, but not how fast it will happen.

14)
Calculate K for the following reaction at 25°C.
2 H2O(l) → 2 H2(g) + O2(g)
K =
ΔG = ΔGprod - ΔGreact = [2(0) + 0] - [2(-237)] = 0 - -474 = 474 kJ
ΔG = - RT ln(K)
474 kJ = - (0.0083145 kJ/K)(298 K) ln(K)
-191.3 = ln(K)  K = 8.3 x 10-84
15)
A state of higher entropy means:
16)
Which of the following would result in a decrease of entropy?
17)
Which of the following proceses must be spontaneous?

A spontaneous reaction has a positive ΔSuniv.

18)
Heat is released during a reaction. This means that:

If heat is released, ΔH is negative, which means that ΔSsurr is positive.

19)
Which of the following conditions would ensure a spontaneous process?

A spontaneous reaction has a positive ΔSuniv.

20)
Which of the following processes would definitely be spontaneous?

A spontaneous reaction has a positive ΔSuniv.

21)
Which would be the most spontaneous at higher temperatures?
22)
Which temperature would allow the following process to be spontaneous?
ΔS = 30 J/K ; ΔH = 120 kJ

Since both ΔH & ΔS are positive, the reaction will be spontaneous at higher temperatures. We can find the temperature at equilibrium (where ΔG = 0) and any temperature above that, there will be a spontaneous reaction.

ΔG = ΔH - TΔS 0 = 120 - (T)(0.03) T = 4000 K
23)
Calculate standard free energy for a reaction that has K = 1.47 x 109. [T = 298 K]
ΔG = - RT ln(K)
ΔG = - (0.0083145 kJ/K)(298 K) ln(1.47 x 109) = -52 kJ
24)
Which of the following processes is the fastest?
25)
Which reaction is most spontaneous (most product produced)?

Choice a & b are both spontaneous, but b is more spontaneous because the number is more negative.

26)
Which reaction is most spontaneous (most product produced)?

The larger the K, the more product produced.