AP‌‌Chemistry‌‌Unit‌‌6‌‌Quiz‌
1.‌‌Refer‌‌to‌‌the‌‌diagram‌‌below.‌‌Which‌‌of‌‌the‌‌following‌‌best‌‌describes‌‌the‌‌reaction?‌
A. Exothermic,‌‌it‌‌gives‌‌out‌‌heat‌
B. Endothermic,‌‌it‌‌gives‌‌out‌‌heat‌
C. Exothermic,‌‌it‌‌absorbs‌‌heat‌
D. Endothermic,‌‌it‌‌absorbs‌‌heat‌
2.‌‌Refer‌‌to‌‌the‌‌diagram‌‌below‌‌(may‌‌not‌‌be‌‌drawn‌‌exactly‌‌to‌‌scale)‌
What‌‌is‌‌the‌‌activation‌‌energy‌‌for‌‌the‌‌reverse‌‌reaction?‌
A. 80‌‌kJ‌
B. 60‌‌kJ‌
C. 40‌‌kJ‌
D. 20‌‌kJ‌
3.‌‌A‌‌100‌‌g‌‌metal‌‌block‌‌with‌‌the‌‌specific‌‌heat‌‌capacity‌‌of‌‌480‌‌J‌‌kg‌-1‌‌ ‌K-1‌
‌ ‌ ‌is‌‌heated‌‌in‌‌a‌‌beaker‌‌of‌
water‌‌until‌‌it‌‌is‌‌boiling‌‌at‌‌room‌‌temperature‌‌(20°C).‌‌After‌‌that,‌‌the‌‌metal‌‌block‌‌is‌‌immediately‌
transferred‌‌into‌‌a‌‌cooler‌‌beaker‌‌of‌‌water.‌‌The‌‌temperature‌‌rise‌‌in‌‌the‌‌cooler‌‌beaker‌‌of‌‌water‌‌is‌‌1°C.‌
What‌‌is‌‌the‌‌mass‌‌of‌‌cold‌‌water?‌‌Assume‌‌no‌‌energy‌‌loss‌‌to‌‌surroundings.‌
(Specific‌‌heat‌‌capacity‌‌of‌‌water‌‌is‌‌4200‌‌J‌‌kg‌-1‌‌ ‌K-1‌
‌ )‌
A. 0.142‌‌kg‌
B. 0.914‌‌kg‌
C. 1.14‌‌kg‌
D. The‌‌question‌‌cannot‌‌be‌‌solved‌‌as‌‌the‌‌final‌‌temperature‌‌is‌‌not‌‌given.‌
4.‌‌A‌‌wooden‌‌block‌‌and‌‌a‌‌metal‌‌block‌‌are‌‌heated‌‌up‌‌at‌‌the‌‌same‌‌time‌‌at‌‌the‌‌same‌‌rate.‌‌The‌‌metal‌
block‌‌is‌‌twice‌‌the‌‌mass‌‌of‌‌the‌‌wooden‌‌block.‌‌The‌‌wooden‌‌block‌‌has‌‌a‌‌specific‌‌heat‌‌capacity‌‌3‌
times‌‌of‌‌the‌‌metal‌‌block.‌‌The‌‌temperature‌‌of‌‌the‌‌metal‌‌block‌‌is‌‌increased‌‌by‌‌k°C while‌‌the‌‌
temperature‌‌of‌‌the‌‌wooden‌‌block‌‌is‌‌increased‌‌by‌‌n°C .‌‌What‌‌is‌ nk ?‌ ‌
A.
3
2
‌
B.
2
3
‌
C.
1
2
‌
D.
1
6
‌
5.‌‌Which‌‌of‌‌the‌‌following‌‌processes‌‌release‌‌the‌‌most‌‌amount‌‌of‌‌energy?‌
A. 50‌‌g‌‌of‌‌ice‌‌melting‌
B. 50‌‌g‌‌of‌‌metal‌‌cooling‌‌of‌‌500‌‌J‌‌kg‌-1‌‌ ‌K-1‌
‌ ‌ ‌down‌‌from‌‌70‌°C to‌‌20‌°C ‌
C. 50‌‌g‌‌of‌‌steam‌‌condensing‌‌into‌‌water‌
D. 50‌‌g‌‌of‌‌water‌‌freezing‌‌into‌‌ice‌
6.‌‌Maya‌‌wants‌‌to‌‌warm‌‌up‌‌a‌‌glass‌‌of‌‌milk‌‌(250mL)‌‌which‌‌is‌‌4°C .‌‌About‌‌how‌‌much‌‌energy‌‌has‌‌to‌‌
be‌‌supplied‌‌to‌‌warm‌‌up‌‌her‌‌glass‌‌of‌‌milk‌‌to‌‌40‌°C ?‌‌The‌‌density‌‌of‌‌milk‌‌is‌‌1.035‌‌kg/L‌‌while‌‌the‌‌
specific‌‌heat‌‌capacity‌‌is‌‌3930‌‌J‌‌kg‌-1‌‌ ‌K-1‌
‌
A. 9315‌‌J‌
B. 34200‌‌J‌
C. 36600‌‌J‌
D. 354000‌‌J‌
7.‌‌The‌‌heat‌‌capacity‌‌of‌‌0.5kg‌‌of‌‌carrots‌‌is‌‌940‌‌J‌‌K-1‌
‌ ‌,‌‌the‌‌heat‌‌capacity‌‌of‌‌1.5‌‌kg‌‌of‌‌cabbage‌‌is‌‌2940‌
J‌‌K-1‌
‌ ‌while‌‌the‌‌heat‌‌capacity‌‌of‌‌3‌‌kg‌‌of‌‌beef‌‌is‌‌5010‌‌J‌‌K-1‌
‌ .‌‌The‌‌food‌‌are‌‌in‌‌room‌‌temperature,‌‌and‌
must‌‌be‌‌heated‌‌up‌‌to‌‌100‌°C ‌to‌‌be‌‌safe‌‌to‌‌eat.‌‌Which‌‌of‌‌the‌‌following‌‌combinations‌‌require‌‌the‌‌
least‌‌amount‌‌of‌‌energy‌‌for‌‌cooking?‌
A. 1.7‌‌kg‌‌of‌‌carrots,‌‌2‌‌kg‌‌of‌‌beef‌‌and‌‌3.1‌‌kg‌‌of‌‌cabbage‌
B. 2‌‌kg‌‌of‌‌carrots,‌‌1.2‌‌kg‌‌of‌‌beef‌‌and‌‌2.4‌‌kg‌‌of‌‌cabbage‌
C. 4.3‌‌kg‌‌of‌‌carrots,‌‌0.4‌‌kg‌‌of‌‌beef,‌‌1.9‌‌kg‌‌of‌‌cabbage‌
D. 1.9‌‌kg‌‌of‌‌carrots,‌‌2.9‌‌kg‌‌of‌‌beef,‌‌1.9‌‌kg‌‌of‌‌cabbage‌
8.‌‌What‌‌is‌‌the‌‌change‌‌in‌‌enthalpy‌‌of‌‌the‌‌below‌‌reaction?‌
AB 2 + 3C −−> D + 2EF ‌
Substance‌
ΔH ° ‌
AB 2 ‌
1.79‌‌kJ/mol‌
EF ‌
-0.57‌‌kJ/mol‌
A. 2930‌‌J‌
B. -2930‌‌J‌
C. -3660‌‌J‌
D. The‌‌answer‌‌can‌‌not‌‌be‌‌determined‌‌as‌‌we‌‌do‌‌not‌‌know‌‌the‌‌standard‌‌enthalpy‌‌of‌‌C‌‌and‌‌D‌
9.‌‌The‌‌following‌‌reaction‌‌occurs:‌
C H 3 OH + 128.49 kJ → 2 H 2 + CO ‌
If‌‌100.8g‌‌of‌‌H 2 is‌‌produced,‌‌how‌‌many‌‌kJ‌‌of‌‌heat‌‌will‌‌be‌‌absorbed?‌ ‌
A. 12951‌‌kJ‌
B. 6475.9‌‌kJ‌
C. 6424.5‌‌kJ‌
D. 7844.9‌‌kJ‌
10.‌‌Carbon‌‌and‌‌hydrogen‌‌are‌‌combined‌‌to‌‌form‌‌methane‌‌in‌‌a‌‌closed‌‌container.‌
‌C (s) + H 2 (g) → CH 4 (g)
ΔH = − 74.85 kJ ‌
If‌‌30‌‌g‌‌of‌‌H 2 ‌is‌‌used,‌‌how‌‌much‌‌heat‌‌will‌‌be‌‌released?‌ ‌
A. 2246‌‌kJ‌
B. 2228‌‌kJ‌
C. 1123‌‌kJ‌
D. 1114‌‌kJ‌
11.‌‌A‌‌copper‌‌block‌‌is‌‌being‌‌heated‌‌up‌‌in‌‌room‌‌conditions.‌‌The‌‌result‌‌as‌‌follows:‌
Mass‌‌of‌‌copper‌‌block‌
m‌‌kg‌
Final‌‌temperature‌
45‌°C ‌
Initial‌‌joulemeter‌‌reading‌
R1 J‌ ‌
Finial‌‌joulemeter‌‌reading‌
R2 J‌ ‌
What‌‌is‌‌the‌‌specific‌‌heat‌‌capacity‌‌of‌‌the‌‌copper‌‌block?‌
A. 26 m (R2 − R1 ) ‌
B.
R2 −R1
15m
‌
C.
R1 −R2
15m
‌
D. The‌‌specific‌‌heat‌‌capacity‌‌can‌‌not‌‌be‌‌determined‌‌because‌‌the‌‌initial‌‌temperature‌‌is‌
unknown‌
12.‌‌Two‌‌solids‌‌X‌‌and‌‌Y‌‌of‌‌equal‌‌mass‌‌are‌‌separately‌‌heated‌‌by‌‌two‌‌identical‌‌heaters.‌‌The‌‌graph‌
below‌‌shows‌‌the‌‌variation‌‌of‌‌the‌‌temperatures‌‌of‌‌the‌‌substances‌‌with‌‌time.‌‌Which‌‌of‌‌the‌‌following‌
statements‌‌is/are‌‌false?‌
I.
II.
The‌‌melting‌‌point‌‌of‌‌X‌‌is‌‌higher‌‌than‌‌that‌‌of‌‌Y.‌
The‌‌specific‌‌heat‌‌capacity‌‌of‌‌X‌‌is‌‌larger‌‌than‌‌that‌‌of‌‌Y.‌
A. I‌‌only‌
B. II‌‌only‌
C. I‌‌and‌‌II‌‌only‌
D. None‌‌of‌‌the‌‌above‌
13.‌ ‌Given‌‌that‌
Reaction‌
Enthalpy‌‌(‌Δ H°)‌ ‌
A‌‌+‌‌B‌2‌ ‌→‌‌AB‌2‌
+283‌‌kJ‌
2AB‌2‌‌ ‌→‌‌2AB‌‌+‌‌B‌2‌
-111‌‌kJ‌
Find‌‌Δ H‌‌of‌‌2A‌‌+‌‌B‌2‌‌ ‌→‌‌2AB‌ ‌
A. 172‌‌kJ‌
B. 394‌‌kJ‌
C. 455‌‌kJ‌
D. 505‌‌kJ‌
14.‌‌Given‌‌that‌
Reaction‌
Enthalpy‌‌(‌Δ H°)‌ ‌
2AB‌‌→‌‌A‌2‌‌ ‌+‌‌B‌2‌
108‌‌kJ‌
A‌2‌‌ ‌+‌‌4C‌‌→‌‌2AC‌2‌
-366‌‌kJ‌
Find‌‌Δ H‌‌of‌‌2AB‌‌+‌‌4C‌‌→‌‌2AC‌2‌‌ ‌+‌‌B‌2‌ ‌
A. -258‌‌kJ‌
B. 258‌‌kJ‌
C. -474‌‌kJ‌
D. 474‌‌kJ‌
15.‌‌Given‌‌that‌
Reaction‌
Enthalpy‌‌(‌Δ H°)‌ ‌
AB‌‌→‌‌A‌‌+‌‌B‌
156‌‌kJ‌
2A‌‌→‌‌A‌2‌
-130‌‌kJ‌
AC‌‌→‌‌A‌‌+‌‌C‌
-277‌‌kJ‌
B‌‌+‌‌C‌‌→‌‌BC‌
-55‌‌kJ‌
Find‌‌Δ H‌‌of‌‌A‌2‌‌ ‌+‌‌BC‌‌→‌‌AB‌‌+‌‌AC‌ ‌
A. -618‌‌kJ‌
B. -306‌‌kJ‌
C. -64‌‌kJ‌
D. 46‌‌kJ‌
16.‌‌Given‌‌that‌
Bond‌
Bond‌‌enthalpy‌
A-A‌
405‌‌kJ‌
B-B‌
365‌‌kJ‌
Find‌‌the‌‌bond‌‌enthalpy‌‌of‌‌A-B‌‌if‌‌the‌‌total‌‌Δ H‌‌of‌‌the‌‌reaction‌‌is‌‌-174‌‌kJ‌ ‌
A. -472‌‌kJ‌
B. 298‌‌kJ‌
C. 385‌‌kJ‌
D. 472‌‌kJ‌
17.‌‌1‌‌kg‌‌of‌‌a‌‌certain‌‌liquid‌‌at‌‌60°C‌‌and‌‌0.5‌‌kg‌‌of‌‌the‌‌same‌‌liquid‌‌at‌‌30°C‌‌are‌‌mixed.‌‌What‌‌is‌‌the‌
final‌‌temperature?‌
A. 42°C‌
B. 45°C‌
C. 50°C‌
D. The‌‌final‌‌temperature‌‌can‌‌not‌‌be‌‌determined‌‌because‌‌the‌‌specific‌‌heat‌‌capacity‌‌is‌‌not‌‌given‌
18.‌‌An‌‌equal‌‌amount‌‌of‌‌energy‌‌is‌‌supplied‌‌to‌‌the‌‌following‌‌substances.‌‌Which‌‌of‌‌the‌‌following‌
will‌‌have‌‌the‌‌smallest‌‌rise‌‌in‌‌temperature?‌
Mass‌‌(kg)‌
Specific‌‌Heat‌‌Capacity‌‌(J‌‌kg‌–1‌‌ ‌K–1‌
‌ )‌
A‌
1‌
4200‌
B‌
2‌
2200‌
C‌
3‌
1400‌
D‌
5‌
830‌
19.‌‌The‌‌graph‌‌below‌‌shows‌‌the‌‌change‌‌in‌‌the‌‌energy‌‌absorbed‌‌by‌‌the‌‌liquids‌‌X,Y‌‌and‌‌Z‌‌of‌‌equal‌
mass‌‌and‌‌their‌‌corresponding‌‌temperature.‌
Let‌‌c‌x‌ ‌,‌‌c‌y‌‌ ‌and‌‌c‌z‌‌ ‌be‌‌the‌‌specific‌‌heat‌‌capacities‌‌of‌‌X,Y‌‌and‌‌Z‌‌respectively.‌‌Which‌‌of‌‌the‌‌following‌
is‌‌correct?‌
A. c‌x‌ ‌=‌‌c‌z‌ ‌>‌‌c‌y‌
B. c‌x‌ =‌
‌ ‌c‌z‌ <‌
‌ ‌c‌y‌
C. c‌y‌ =‌
‌ ‌c‌z‌ <‌
‌ ‌c‌z‌
D. c‌y‌ =‌
‌ ‌c‌z‌ >‌
‌ ‌c‌z‌
20.‌‌Equal‌‌masses‌‌of‌‌four‌‌different‌‌liquids‌‌are‌‌separately‌‌heated‌‌at‌‌the‌‌same‌‌rate.‌‌The‌‌initial‌
temperature‌‌of‌‌all‌‌the‌‌liquids‌‌are‌‌all‌‌20°C.‌‌The‌‌boiling‌‌points‌‌and‌‌specific‌‌heat‌‌capacities‌‌of‌‌the‌
liquids‌‌are‌‌shown‌‌below.‌‌Which‌‌one‌‌of‌‌them‌‌will‌‌boil‌‌first?‌
‌Specific‌‌heat‌‌capacity‌‌(J‌‌kg‌–1‌‌ ‌K–1‌
‌ )‌
Boiling‌‌point‌‌(°C)‌
A‌
1100‌
50‌
B‌
785‌
70‌
C‌
600‌
90‌
D‌
170‌
300‌
Explanations‌
1.‌‌A.‌‌Exothermic‌‌reactions‌‌give‌‌out‌‌heat.‌‌This‌‌can‌‌be‌‌observed‌‌when‌‌the‌‌amount‌‌of‌‌potential‌
energy‌‌decreases.‌‌Energy‌‌can‌‌neither‌‌be‌‌created‌‌nor‌‌destroyed.‌‌Some‌‌of‌‌the‌‌energy‌‌will‌‌be‌
converted‌‌into‌‌heat‌‌energy‌‌and‌‌escape.‌‌B‌‌and‌‌C‌‌are‌‌conceptually‌‌wrong.‌
2.‌‌D‌
Reverse‌‌activation‌‌rate‌‌refers‌‌to‌‌the‌‌difference‌‌in‌‌energy‌‌between‌‌the‌‌product(s)‌‌(right)‌‌and‌‌the‌
transition‌‌state‌‌(hill)‌
80‌‌kJ‌‌-‌‌60‌‌kJ‌‌=‌‌20‌‌kJ‌
3.‌‌B‌
Q = mcΔT ‌
Assume‌‌no‌‌energy‌‌loss‌‌to‌‌surroundings‌
Qlost by metal block = Qgained by water ‌
mmetal cmetal ΔT = mwater cwater ΔT ‌
100‌‌g‌‌=‌‌0.1‌‌kg‌
0.1 × 480 × (100 − 20)= mwater × 4200 × 1
mwater = 0.914 kg (3 sig figs) ‌
4.‌‌B‌
They‌‌are‌‌heated‌‌up‌‌at‌‌the‌‌same‌‌time‌‌at‌‌the‌‌same‌‌rate,‌‌meaning‌‌that‌‌the‌‌energy‌‌supplied‌‌is‌‌the‌‌same.‌
By‌‌the‌‌equation‌‌Q = mcΔT ,‌ ‌
mwood cwood ΔT wood = mmetal cmetal ΔT metal ‌
Let‌‌mass‌‌of‌‌wooden‌‌block‌‌be‌‌m,‌‌mass‌‌of‌‌metal‌‌block‌‌be‌‌2m,‌‌specific‌‌heat‌‌capacity‌‌of‌‌wooden‌
block‌‌be‌‌3c,‌‌specific‌‌heat‌‌capacity‌‌of‌‌metal‌‌block‌‌of‌‌metal‌‌be‌‌c‌
2m × c × n = m × 3c × k ‌
2mc × n = 3mc × k ‌
Cross‌‌out‌‌mc ,‌ ‌
2n = 3k ‌
k
n
=
2
3
‌
5.‌‌B‌
When‌‌steam‌‌condenses‌‌into‌‌water,‌‌more‌‌energy‌‌is‌‌released‌‌than‌‌when‌‌water‌‌freezes‌‌into‌‌ice.‌
Energy‌‌of‌‌particles‌‌in‌‌a‌‌gas‌‌>‌‌energy‌‌of‌‌particles‌‌in‌‌a‌‌liquid‌‌>‌‌energy‌‌of‌ ‌particles‌‌in‌‌a‌‌solid‌
More‌‌energy‌‌is‌‌required‌‌to‌‌turn‌‌gas‌‌into‌‌liquid‌‌than‌‌from‌‌liquid‌‌to‌‌solid‌‌because‌‌more‌‌energy‌‌needs‌
to‌‌be‌‌released‌‌so‌‌that‌‌the‌‌particles.‌
Choice‌‌A‌‌(ice‌‌melting)‌‌takes‌‌up‌‌heat.‌
A‌‌metal‌‌cooling‌‌down‌‌may‌‌release‌‌heat,‌‌but‌‌due‌‌to‌‌the‌‌small‌‌specific‌‌heat‌‌capacity,‌‌the‌‌energy‌
released‌‌is‌‌much‌‌less‌‌than‌‌the‌‌amount‌‌of‌‌energy‌‌released‌‌when‌‌steam‌‌condenses‌‌into‌‌water.‌
6.‌‌C‌
250 mL = 0.25 L ‌
0.25 L ×
1.035 kg
1L
=‌‌0.25875 kg ‌
Q = mcΔT ‌
Q = 0.25875 kg × 3930 J kg −1 K
−1
× (40 − 4)K
Q = 36607.95 J ‌
Q = 36600 J (3 sig figs) ‌
7.‌‌D‌
Convert‌‌the‌‌specific‌‌heat‌‌capacity‌‌of‌‌the‌‌carrots,‌‌cabbage‌‌and‌‌beef‌‌into‌‌the‌‌specific‌‌heat‌‌capacity‌
(1kg)‌‌so‌‌it‌‌is‌‌easier‌‌for‌‌comparison‌
Food‌
Specific‌‌heat‌‌capacity‌‌(J‌‌kg‌-1‌‌ ‌K-1)‌
‌
Carrots‌
1880‌
Beef‌
1960‌
Cabbage‌
1670‌
A:‌‌1.7 × 1880 + 2 × 1960 + 3.1 × 1670 = 12293 ‌
B:‌ 2 × 1880 + 1.2 × 1960 + 2.4 × 1670 = 10120 ‌
C:‌‌4.3 × 1880 + 0.4 × 1960 + 1.9 × 1670 = 12041 ‌
D:‌‌1.9 × 1880 + 2.9 × 1960 + 1.9 × 1670 = 12429 ‌
8.‌‌B‌
ΔS° = ΣS° P roducts − ΣS° Reactants ‌
Standard‌‌enthalpy‌‌of‌‌an‌‌element‌‌(C,‌‌D)‌‌is‌‌0‌
ΔS° = (2 ×− 0.57) − (1.79) = − 2.93 kJ = − 2930 J ‌
‌-2930‌‌J‌
9.‌‌C‌
1 mol H 2
1.008 g H 2
100.8g‌‌H 2 ×
×
128.49 kJ
2 mol H 2
=‌‌6424.5‌‌kJ‌ ‌
10.‌‌D‌
Balance‌‌the‌‌equation‌
C + 2 H 2 → CH 4
‌
1 mol H
kJ
30‌‌g‌‌H 2 × ‌1.008 g H2 × ‌ 274.85
=‌‌1114‌‌kJ‌ ‌
mol H
2
2
11.‌‌B‌
Q = mcΔT ‌
R2 − R1 = mc(45 − 20) ‌
c =‌
R2 −R1
15m
‌
12.‌‌B‌
I‌‌is‌‌true.‌‌The‌‌flat‌‌part‌‌of‌‌the‌‌graph‌‌(plateau)‌‌means‌‌that‌‌the‌‌object‌‌is‌‌changing‌‌state.‌
The‌‌melting‌‌point‌‌of‌‌X‌‌is‌‌higher‌‌than‌‌Y‌‌because‌‌the‌‌corresponding‌‌y‌‌value‌‌(temperature)‌‌is‌‌higher.‌
II‌‌is‌‌false.‌‌A‌‌steeper‌‌slope‌‌means‌‌that‌‌temperature‌‌changes‌‌quicker.‌‌Less‌‌energy‌‌has‌‌to‌‌be‌‌supplied‌
for‌‌a‌‌change‌‌in‌‌state.‌‌The‌‌specific‌‌heat‌‌capacity‌‌of‌‌X‌‌is‌‌smaller‌‌than‌‌that‌‌of‌‌Y.‌
13.‌‌C‌
2A‌‌+‌‌2B‌2‌ ‌→‌‌2AB‌2‌ (Δ H°)‌‌=566‌‌kJ‌ ‌
2AB‌2‌‌ ‌→‌‌2AB‌‌+‌‌B‌2‌ (Δ H°)‌‌=‌‌-111‌‌kJ‌ ‌
2A‌‌+‌‌2B‌2‌ →‌
‌2AB‌‌+‌‌B‌2‌ ‌ (Δ H°)‌‌=455‌‌kJ‌ ‌
‌
2A‌‌+‌‌B‌2‌‌ ‌→‌‌2AB‌‌ (Δ H°)‌‌=455‌‌kJ‌ ‌
14.‌‌A‌
2AB‌‌→‌‌A‌2‌‌ ‌+‌‌B‌2‌
‌
A‌2‌‌ ‌+‌‌4C‌‌→‌‌2AC‌2‌
‌
2AB+‌‌4C‌‌→‌‌2AC‌2‌+‌‌B‌2‌ ‌
Δ H°‌‌=‌‌108‌‌kJ‌ ‌
Δ H°‌‌=‌‌-366‌‌kJ‌ ‌
Δ H°‌‌=‌‌-258‌‌kJ‌ ‌
15.‌‌C‌
Required‌‌reaction:‌ ‌A‌2‌‌ ‌+‌‌BC‌‌→‌‌AB‌‌+‌‌AC‌
A‌2‌‌ ‌→‌ ‌2A‌‌
‌Δ H‌‌=‌‌-130‌‌kJ‌ ‌
A‌‌+‌‌B‌‌→‌‌AB‌
‌Δ H‌‌=‌‌-156‌‌kJ‌ ‌
A‌‌‌+‌‌C‌‌→‌‌AC‌
‌Δ H‌‌=‌‌277‌‌kJ‌ ‌
BC‌‌→‌‌B‌‌+‌‌C‌ ‌
‌ Δ H‌‌=‌‌-55‌‌kJ‌ ‌
Simplify,‌
A‌2‌‌ ‌+‌‌BC‌‌→‌‌AB‌‌+‌‌AC‌‌
Δ H‌‌=‌‌-64‌‌kJ‌ ‌
16.‌‌D‌
Δ H°‌‌=‌‌∑‌Δ H°‌‌products‌‌-‌‌∑‌Δ H°‌‌reactants‌ ‌
Δ H°‌‌=‌‌∑‌Δ H°‌‌bonds‌‌broken‌‌-‌‌∑‌Δ H°‌‌bonds‌‌formed‌ ‌
Let‌‌bond‌‌enthalpy‌‌of‌‌A-B‌‌=‌‌x,‌
-174‌‌=‌ ‌405‌‌+‌‌365‌‌-‌‌2x‌
2x‌‌=‌‌944‌
x‌‌=‌‌472‌‌kJ‌
17.‌‌C‌
Let‌‌the‌‌final‌‌temperature‌‌be‌‌t‌‌and‌‌specific‌‌heat‌‌capacity‌‌be‌‌c,‌‌Q=mcΔT‌
mcΔT=mcΔT‌
1(c)(60-t)=0.5(c)(t-30)‌
Simplify‌‌by‌‌crossing‌‌out‌‌c‌
60‌‌-‌‌t‌‌=‌‌0.5‌‌t‌‌-‌‌15‌
75‌‌=‌‌1.5‌‌t‌
t‌‌=‌‌50‌
18.‌‌B‌
Q=mcΔT.‌‌The‌‌energy‌‌supplied‌‌is‌‌the‌‌same‌
Substance‌‌A:‌‌Q‌‌=‌‌(1)(4200)(ΔT)‌
Q
ΔT‌‌=‌‌ 4200
‌‌
Substance‌‌B:‌‌Q‌‌=‌‌(2)(2200)(ΔT)‌
Q
ΔT‌‌=‌‌ 4400
‌
Substance‌‌C:‌‌Q‌‌=‌‌(3)(1400)(ΔT‌‌)‌
Q
ΔT‌‌=‌‌ 4200
‌
Substance‌‌D:‌‌Q‌‌=‌‌(5)(830)(ΔT)‌
Q
ΔT‌‌=‌‌ 4150
‌
‌The‌‌numerator‌‌is‌‌the‌‌same‌‌(Q).‌‌The‌‌larger‌‌the‌‌denominator,‌‌the‌‌smaller‌‌the‌‌value‌‌of‌‌ΔT.‌‌For‌
substance‌‌B,‌‌the‌‌denominator‌‌is‌‌the‌‌largest‌‌(4400>4200>4150).‌‌Hence,‌‌substance‌‌B‌‌has‌‌the‌
smallest‌‌rise‌‌in‌‌temperature.‌
19.‌‌A‌
in y
Q
Q=mcΔT,‌‌Slope‌‌=‌‌ change
=‌‌ ΔT
=‌‌mc‌ ‌
change in x
The‌‌mass‌‌is‌‌the‌‌same.‌‌Therefore,‌‌the‌‌steeper‌‌the‌‌slope,‌‌the‌‌greater‌‌the‌‌specific‌‌heat‌‌capacity.‌
Slope‌‌of‌‌X‌‌=‌‌slope‌‌of‌‌Z.‌‌c‌x‌ =‌
‌ ‌c‌z‌.‌‌Slope‌‌of‌‌Y ‌‌is‌‌smaller‌‌than‌‌X .‌‌mc‌x‌ >‌
‌ ‌mc‌y‌‌ ‌,‌‌m‌‌is‌‌the‌‌same,‌‌so‌‌c‌x‌ >‌
‌ ‌c‌y‌
Hence,‌‌c‌x‌ =‌
‌ ‌c‌z‌ >‌
‌ ‌c‌y‌
20.‌‌A‌
Q=mcΔT‌
Liquid‌‌A:‌‌Q‌‌=‌‌(m)(1100)(50-20)‌‌=‌‌33000m‌
Liquid‌‌B:‌‌Q‌‌=‌‌(m)(785)(70-20)‌‌=39250‌‌m‌
Liquid‌‌C:‌‌Q=(m)(600)(90-20)=42000m‌
Liquid‌‌D:‌‌Q=(m)(170)(300-20)=47600m‌
The‌‌liquids‌‌have‌‌the‌‌same‌‌mass,‌‌so‌‌m‌‌is‌‌the‌‌same‌
The‌‌energy‌‌is‌‌supplied‌‌at‌‌the‌‌same‌‌rate.‌‌The‌‌first‌‌liquid‌‌to‌‌reach‌‌the‌‌required‌‌amount‌‌of‌‌energy‌‌to‌
boil‌‌will‌‌boil‌‌first.‌‌33000‌‌<‌‌39250‌‌<‌‌42000‌‌<‌‌47600,‌‌so‌‌liquid‌‌A‌‌requires‌‌the‌‌least‌‌amount‌‌of‌
energy‌‌to‌‌reach‌‌the‌‌boiling‌‌point.‌‌Liquid‌‌A‌‌will‌‌boil‌‌first.‌