Solutions to Unit 4 Homework – Chemistry 1A
1. When brick 1 is brought in contact
with brick 2, heat flows from brick 1 to
brick 2, until thermal equilibrium is
reached. This equilibrium occurs when
the final temperature of the two bricks
become 50oC. The heat flowing out of
brick 1 equals the heat absorbed by
brick 2.
INTIAL
FINAL
Brick 1
Brick 2
T = 100oC
m = 5 kg
T = 5oC
m=?
Brick 1
T = 50oC
m = 5 kg
Brick 2
T = 50oC
m=?
Heat delivered by brick 1 is


q   CdT  mC o T2  T1   5kg  800 JK 1kg 1 323K  373K   200,000 J
This is the heat absorbed by brick 2. Therefore, for brick 2,


 200,000 J  mC o T2  T1   m  900 JK 1kg 1 323K  273K 
m 
200,000 J
 4.93kg
323K  278K  900 JK 1kg 1


4.15
4.1
2. Using the Van’t Hoff equation, a plot of lnK versus
1/T will yield the slope -rHo/R.
ln
1 1
  
 T2 T1 
4.05
LnK
K P T2 
 H
 r
K P T1 
R
o
y = 1505.9x + 1.8538
R2 = 0.9994
4
3.95
3.9
3.85
In this case slope = 1505.9, or ΔH = -12.52 KJmol-1.
3.8
0.00125 0.0013 0.00135 0.0014 0.00145 0.0015 0.00155
1/T (K-1)
3. (a)
(b)
4/5/6 Are more or less the same kinds of problems.
7. We have the following information:
vapHo at 373 K is 40.7 kJmol-1
C p (l) = 75.2 Jmol-1K-1
C p (g) = 33.6 Jmol-1K-1
H2O (l)
Ho(373K)
H2O (g)
H2
H1
H2O (l)
Ho(298K)
H2O (g)
o
1
2
o
 vap H 298
K  H  H  H 373K

373K
298K
C PL dT  40.7kJmol 1  
298K
373K
C PG dT  75.2373  298  40.7  10 3  33.6298  373
 43.82kJmol 1
8. H T2   H T1   C P T2  T1   520  12323  343  280kJmol1
9. (a) Using Hess’s law:
C6H12O6(s)
ΔHro
2 CH3CH(OH)COOH(s)
2(- 1344) kJ mol-1
= - 2688 kJ mol-1
- 2808 kJ mol-1
6 CO2(g) + 6 H2O(l)

 2808kJmol 1  H ro   2688kJmol 1

 H ro  2808  2688  120kJmol 1
It follows that the standard enthalpy for the conversion of glucose to lactic acid during
glycolysis is -120 kJ mol-1, about 4 % of the enthalpy change of combustion of glucose.
Therefore, full oxidation of glucose is metabolically more useful than glycolysis, because in
the former process more energy becomes available for performing work.
(b) Using Hess’s law:
C6H12O6(s)
ΔHro
2 C2H5OH(l) + 2 CO2
(g)
[+ 6O2(g)]
2( -1368 kJ mol-1)
= - 2736 kJ mol-1
- 2808 kJ mol-1
6 CO2(g) + 6 H2O(l)

 2808kJmol 1  H ro   2736kJmol 1

 H ro  2808  2736  72kJmol 1
10. The Kekulē structure of benzene has 3 C=C bonds and 3 C-C bonds. The total bond
energy corresponding to the ring structure is 3(348) + 3(612) = 2880 kJ mol-1. Compare
this to the resonance structure which has 6 aromatic C-C bonds, giving 6(518) = 3108 kJ
mol-1. Thus, more energy is required to break the resonance form of benzene. The
resonance form is therefore more stable than the corresponding Kekulē structure, by an
amount 228 kJ mol-1.
11.
12
13
(a) The Mg2+ + ATP complexation can be studied in an electrochemical cell. We want to
determine the activity (concentration) of Mg2+ at any concentration in the presence of 0.100 M
ATP.
Putting ATP in a solution of known Mg2+(aq) will result in the consumption of some of the
Mg2+, thus decreasing it to some lower value. We need to determine this new value.
Construct the following concentration cell.
(see next page).
V
Salt bridge
Mg electrode
Mg electrode
2+
Mg2+
/ATM
Mg
/ATP
ANODE
[Mg2+] = 1M
CATHODE
Mg (s)  Mg2+ (x M) + 2e
Mixture of Mg2+ and ATP
Mg2+ (1 M) + 2e-  Mg (s)
Mg2+/Mg Standard conditions
The overall reaction is: Mg 2+ (1M)  Mg2+ (x M) with Eocell = 0V, and n = 2.
The reaction quotient is Q 
a Mg 2 
1M
 a Mg 2 
The electrodes are identical, Eo are numerically equal. Therefore, Eocell = 0V.
However, the Mg2+ ion concentrations differ in each half-cell. This results in a potential
difference that can be measured using the voltmeter. If the two half-cells were at equilibrium,
then the Mg2+ concentrations would be identical, and Ecell = 0. However, this is not the case.
This is a concentration cell, and the value of Ecell can be used to determine the unknown
concentration of Mg2+ in the LHS cell.
The Nernst equation for the above electrochemical cell is:
E cell  E ocell 
E cell  
RT
RT
ln Q  0 
ln a Mg 2
nF
2F
RT
ln a Mg 2
2F
By measuring the potential difference, we can determine the activity of Mg2+ at any
concentration in the presence of ATP.
(b) If ATP reacts with Mg2+, the new equilibrium concentrations will be:
[Mg2+]eq = x – y
[ATP]eq = 0.100 – y
[complex]eq = y
The equilibrium constant is given by K 
y
x  y0.100  y
Here, x refers to the initial concentration of Mg2+ which we know. The value x – y is simply the
concentration of Mg2+ at equilibrium, which we have established in (a) (the activity).
Therefore we can determine y and hence the value of K.
Note, although our reaction between Mg2+ and ATP reaches equilibrium in the LHS cell,
the two half cells are not at equilibrium with respect to each other because the Mg2+
concentrations differ. If they were, we would not measure a potential difference between
them, implying that the concentration of Mg2+ in the LSH cell must me 1M.