Exercício Resolvido - Quimica Geral (137)

Prévia do material em texto

Chapter 7: Thermochemistry 
 273 
20. (M) The amounts of the two reactants provided are the same as their stoichiometric 
coefficients in the balanced equation. Thus 852 kJ of heat is given off by the reaction. We 
can use this quantity of heat, along with the specific heat of the mixture, to determine the 
temperature change that will occur if all of the heat is retained in the reaction mixture. 
We make use of the fact that mass sp.ht. T . 
5 2 3
2 3
2 3
102 g Al O 55.8 g Fe 0.8 J
heat = 8.52 10 J = 1 mol Al O + 2 mol Fe 
1 mol Al O 1 mol Fe g C
T 
5
3
1 1
8.52 10 J
 = = 5 10 C 
214 g 0.8 J g C
T 
The temperature needs to increase from 25 C to 1530 C or 3 = 1505 C = 1.5 10 CT . 
Since the actual T is more than three times as large as this value, the iron indeed will 
melt, even if a large fraction of the heat evolved is lost to the surroundings and is not 
retained in the products. 
 
21. (M) 
 (a) We first compute the heat produced by this reaction, then determine the value of H in 
kJ/mol KOH. 
 2calorimeter rxn
J
 = 0.205 + 55.9 g 4.18 24.4 C 23.5 C = 2 10 J heat = 
g C
q q 
 H
2 10
0 205
5 10
2
1
J
1 kJ
1000J
g
1mol KOH
56.1g KOH
kJ / mol
.
 
 
 (b) The here is known to just one significant figure (0.9 C). Doubling the amount 
of KOH should give a temperature change known to two significant figures (1.6 C) 
and using twenty times the mass of KOH should give a temperature change known 
to three significant figures (16.0 C). This would require 4.10 g KOH rather than the 
0.205 g KOH actually used, and would increase the precision from one part in five 
to one part in 500, or 0.2 %. Note that as the mass of KOH is increased and the 
mass of H2O stays constant, the assumption of a constant specific heat becomes less 
valid. 
 
22. (M) First we must determine the heat absorbed by the solute during the chemical reaction, 
qrxn. This is the negative of the heat lost by the solution, qsoln. Since the solution (water plus 
solute) actually gives up heat, the temperature of the solution drops. 
rxn
1 L 2.50 mol KI 20.3 kJ
heat of reaction = 150.0 mL × × × = 7.61 kJ = 
1000 mL 1 L soln 1 mol KI
q 
rxn soln
1.30 g 2.7 J
= = 150.0 mL 
1 mL g C
q q T 
3
7.61 10 J
14 C
1.30g 2.7 J
150.0 mL
1mL g C
T 
final = initial + = 23.5 C 14 C = 10. C T T T 
 
Chapter 7: Thermochemistry 
274 
23. (M) Let x be the mass, (in grams), of NH Cl4 added to the water. 
heat = mass sp.ht. T 
4
44
1 mol NH Cl 14.7 kJ 1000 J
 × × ×
53.49 g NH Cl 1 mol NH Cl 1 kJ
x
1.00 g J
= 1400 mL + 4.18 10. 25 C 
1 mL g C
x 
4
4 2
4
8.8 10
275 = 8.8 10 + 63 ; = = 4.2 10 g NH Cl 
275 63
x x x 
Our final value is approximate because of the assumed density (1.00 g/mL). The solution’s 
density probably is a bit larger than 1.00 g/mL. Many aqueous solutions are somewhat more 
dense than water. 
 
24. (M) heat 
aOH
H
kJ= 500 mL 
1 L
1000 mL
 
7.0 mol N
1 L soln
 
44.5 kJ
1 mol NaO
 = 1.6 102 
= heat of r = heat abso =eaction rbed by solution OR rxn solnq q 
51.6 10 J
74 C
1.08g 4.00J
500.mL
1mL g C
T final 21 C +74 C = 95 CT 
 
25. (E) We assume that the solution volumes are additive; that is, that 200.0 mL of solution is 
formed. Then we compute the heat needed to warm the solution and the cup, and finally 
 H for the reaction. 
31.02 g J Jheat = 200.0 mL 4.02 27.8 C 21.1 C + 10 27.8 C 21.1 C = 5.6 10 J 
1 mL g C C
 
3
neutr
5.6 10 J 1 kJ
 = = 56 kJ/mol 55.6 kJ/mol to three significant figures 
0.100 mol 1000 J
H 
 
26. (M) Neutralization reaction: 2NaOH aq + HCl aq NaCl aq + H O(l) 
Since NaOH and HCl react in a one-to-one molar ratio, and since there is twice the volume 
of NaOH solution as HCl solution, but the [HCl] is not twice the [NaOH], the HCl solution 
is the limiting reagent. 
2
2
rxn soln
1 mol H O1 L 1.86 mol HCl 55.84 kJ
heat released = 25.00 mL = 2.60 kJ 
1000 mL 1 L 1 mol HCl 1 mol H O
= heat of reaction = heat absorbed by solution or = q q
 
32.60 10 J
8.54 C
1.02g 3.98J
75.00mL
1mL g C
T T = Tfinal Ti Tfinal = T + Ti 
8.54 C + 24.72 C =33.26 CfinalT