Score = /22 = %
KEY
1. 20 mL of 0.1 M H
3
PO
4
and 20 mL of 0.3 M NaOH are combined in a coffee cup calorimeter. The temperature of the solution increases by 1.90°C when the following reaction occurs.
H
3
PO
4
(aq) + 3NaOH(aq)  Na
3
PO
4
(aq) + 3H
2
O(l)
The specific heat of the solution is 4.18. What is the ΔH of the reaction? (6 pts)
First we find the heat gained by the solution in kilojoules: q = mcΔT = 40(4.18)(1.90) = 317.7 J = 0.3177 kJ
We must also know number of moles that reacted:
20 mL = 0.020 L
Then calculate ΔH:
0.020(0.1) = 0.002 mol of H q = nΔH
-0.3177 = 0.002(ΔH)
ΔH = -158.8 kJ/mol
2. Find the ΔH for the reaction below. (6 pts)
(g)
3
PO
4
CH
4
(g) + NH
3
(g)  HCN(g) + 3H
2
Use the following reactions and their ΔH values:
N
2
(g) + 3H
2
(g) → 2NH
3
(g) ∆H = -91.8 kJ/mol
C(s) + 2H
2
(g) → CH
4
(g) ∆H = -74.9 kJ/mol
H
2
(g) + 2C(s) + N
2
(g) → 2HCN(g) ∆H = +270.3 kJ/mol
CH
4
(g)  C(s) + 2H
2
(g)
NH
3
(g)  1/2 N
2
(g) + 3/2 H
1/2 H
2
(g) + C(s) + 1/2 N
2
(g) → HCN(g)
2
∆H = +74.9
(g) ∆H = +91.8/2
∆H = +270.3/2
ΔH = +255.95 kJ/mol

3. Given the following Lewis structures and bond enthalpy values:
Bond Bond Enthalpy (kJ/mol) Bond Bond Enthalpy (kJ/mol)
C-C 347
C=C 598
C  C 813
C-N 285
C=N 616
C-H 413
C=O 805
O=O 498
H-O
H-Cl
464
431
C  N 866
C-Br 285
N=N 418
N  N 946
H-N
H-H
391
436
Br-Br 193
Cl-Cl 242
N-O 222 N=O 590 a.
Calculate the ΔH of the following reaction: (6 pts)
CH
4
(g) + NH
3
(g)  HCN(g) + 3H
2
(g)
Sum of the bond enthalpies of the reactants (energy in) = 4(413) + 3(391) = 2825
Sum of the bond enthalpies of the products (energy out)= 413 + 866 + 3(436) = 2587
ΔH = 2825 – 2587 = +238 kJ/mol b.
For the reaction in part a, would the ΔS be positive or negative? Explain your answer.
(2 pts)
Positive. The products consist of 4 moles of gas, the reactants only 2. When more gas is produced, entropy increases. c.
Based on your answers to parts a and b, would you expect the reaction to be: (2 pts) a.
always spontaneous b.
never spontaneous c.
spontaneous if the temperature is high enough d.
spontaneous if the temperature is low enough
Explain your answer.
The positive ΔH favors a nonspontaneous reaction, but the positive ΔS favors a spontaneous reaction. To make the TΔS term outweigh the ΔH, the temperature must be high.