Grade 12 High School - General Chemistry 2 Notes

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CHAPTER‌‌1:‌‌THERMOCHEMISTRY‌ ‌ Specific‌‌Heat‌‌Capacity‌‌‌(c)‌ ‌
‌ Quantity‌‌of‌‌heat‌‌required‌‌to‌‌change‌‌the‌‌temperature‌‌of‌‌1‌‌gram‌‌of‌‌a‌‌
MEETING‌‌#1‌ ‌ substance‌‌by‌‌1‌‌K.‌ ‌
‌ Unit‌‌is‌‌J/g‌‌K.‌ ‌
Thermochemistry‌ ‌ ‌
Branch‌‌of‌‌thermodynamics‌‌that‌‌deals‌‌with‌‌the‌‌heat‌‌absorbed‌‌or‌‌produced‌‌by‌‌ q = mcΔT ‌
a‌‌chemical‌‌reaction.‌ ‌ Molar‌‌Heat‌‌Capacity‌‌‌(c‌n‌)‌ ‌
‌ Quantity‌‌of‌‌heat‌‌required‌‌to‌‌change‌‌the‌‌temperature‌‌of‌‌1‌‌mole‌‌of‌‌
Exothermic‌ ‌ substance‌‌1‌‌K.‌ ‌
Releases‌‌energy.‌ ‌ Unit‌‌is‌‌J/mol‌‌K.‌ ‌
ex.‌‌Cellular‌‌respiration,‌‌freezing‌‌(liquid‌‌to‌‌solid),‌‌condensation‌‌(gas‌‌to‌‌ ‌
liquid),‌‌deposition‌‌(gas‌‌to‌‌solid),‌‌combination‌‌and‌‌combustion‌‌reactions‌ ‌ q = ncn ΔT ‌
‌ ‌
Endothermic‌ ‌ Thermal‌‌Equilibrium‌
Absorbs‌‌energy.‌ ‌ Two‌‌bodies‌‌initially‌‌at‌‌different‌‌temperatures‌‌will‌‌eventually‌‌reach‌‌
ex.‌‌Photosynthesis,‌‌melting‌‌(solid‌‌to‌‌liquid),‌‌evaporating‌‌(liquid‌‌to‌‌gas),‌‌ equilibrium‌‌at‌‌the‌‌same‌‌temperature.‌ ‌
sublimation‌‌(solid‌‌to‌‌gas),‌‌decomposition‌‌reactions‌ ‌ ‌
‌ −q =q ‌
Heat‌‌‌vs.‌E‌ nergy‌ ‌ ‌
‌ 1st‌‌Law‌‌of‌‌Thermodynamics‌‌‌(Law‌‌of‌‌Conservation‌‌of‌‌Energy)‌ ‌
Heat‌‌‌(q)‌ ‌ Energy,‌‌including‌‌thermal‌‌energy‌‌(heat),‌‌can‌‌neither‌‌be‌‌created‌‌nor‌‌
A‌‌continuous‌‌flow‌‌of‌‌energy‌‌from‌‌high‌‌temperature‌‌(which‌‌has‌‌a‌‌high‌‌ destroyed.‌ ‌‌
amount‌‌of‌‌KE)‌‌to‌‌low‌‌temperature‌‌(low‌‌KE).‌ ‌
Results‌‌in‌‌a‌‌change‌‌of‌‌temperature.‌ ‌
As‌‌an‌‌energy,‌‌it‌‌is‌‌called‌‌thermal‌‌energy.‌ ‌
‌
Energy‌ ‌
Ability‌‌to‌‌do‌‌work.‌ ‌
ex.‌‌Steam‌‌engine‌‌(heat‌‌doing‌‌work)‌ ‌
‌
W = F dcosθ ‌
‌
System‌ ‌
The‌‌item‌‌or‌‌reaction‌‌of‌‌interest.‌ ‌
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Open‌‌System‌ ‌
Allows‌‌exchange‌‌of‌‌energy‌‌and‌‌matter.‌ ‌
ex.‌‌Puddle‌‌in‌‌the‌‌sun,‌‌water‌‌evaporates‌‌and‌‌KE‌‌is‌‌transferred‌ ‌
‌
Closed‌‌System‌ ‌
Allows‌‌exchange‌‌of‌‌energy‌‌but‌‌not‌‌matter.‌ ‌
ex.‌‌Simulation‌‌with‌‌iron‌‌and‌‌brick‌‌(VLQ‌‌1)‌ ‌
‌
Isolated‌‌System‌ ‌
Does‌‌not‌‌allow‌‌exchange‌‌of‌‌energy‌‌and‌‌matter.‌ ‌
ex.‌‌Thermos‌ ‌
‌
Surrounding‌ ‌
Everything‌‌else‌‌that‌‌is‌‌not‌‌the‌‌system.‌ ‌
‌
Heat‌‌Transfer‌‌‌and‌T‌ emperature‌‌Changes‌ ‌
‌
Calorimetry‌ ‌
The‌‌measure‌‌of‌‌heat‌‌flow.‌ ‌
Uses‌‌calorimeter.‌ ‌
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Heat‌‌Capacity‌‌‌(C)‌ ‌
Quantity‌‌of‌‌heat‌‌required‌‌to‌‌change‌‌its‌‌temperature‌‌by‌‌1‌‌K.‌ ‌
Unit‌‌is‌‌J/K.‌ ‌
‌
q = C ΔT ‌
‌
If‌‌q‌‌>‌‌0,‌‌then‌‌it‌‌is‌‌endothermic,‌‌absorbs‌‌heat.‌ ‌
If‌‌q‌‌<‌‌0,‌‌then‌‌it‌‌is‌‌exothermic,‌‌releases‌‌heat.‌ ‌
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Low‌‌Heat‌‌Capacity‌ ‌
Objects‌‌that‌‌can‌‌easily‌‌heat‌‌up.‌ ‌
ex.‌‌Metals,‌‌conductors‌ ‌
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‌

, ‌
HEAT‌‌TRANSFER‌a‌ nd‌‌TEMPERATURE‌‌CHANGES‌ ‌ 3. How‌‌much‌‌heat‌‌is‌‌needed‌‌to‌‌warm‌‌2.75‌‌mol‌‌of‌‌water‌‌
‌ from‌‌22.0‌‌°C‌‌to‌‌near‌‌its‌‌boiling‌‌point,‌‌98‌‌°C?‌‌The‌‌
Heat‌‌Capacity‌ q = C ΔT ‌ specific‌‌heat‌‌of‌‌water‌‌is‌‌4.18‌‌J/g‌‌K.‌ ‌
Specific‌‌Heat‌‌Capacity‌ q = mcΔT ‌ ‌
Molar‌‌Heat‌‌Capacity‌ q = nC n ΔT ‌ G:‌ n‌‌=‌‌2.75‌‌mol‌ R:‌‌q‌ ‌
‌ T‌i‌‌ ‌=‌‌22.0‌‌°C‌ ‌
‌ T‌f‌‌ ‌=‌‌98‌‌°C‌ ‌
1. Large‌‌bed‌‌of‌‌rocks‌‌are‌‌used‌‌in‌‌some‌‌solar-heated‌‌ c‌‌=‌‌4.18‌‌J/g‌‌°C‌ ‌
homes‌‌to‌‌store‌‌heat.‌‌Calculate‌‌the‌‌quantity‌‌of‌‌heat‌‌ ‌
absorbed‌‌by‌‌50.0‌‌kg‌‌of‌‌rocks‌‌if‌‌their‌‌temperature‌‌ E:‌ q = mcΔT ‌
increases‌‌by‌‌12.0‌‌°C.‌‌Assume‌‌that‌‌the‌‌specific‌‌heat‌‌of‌‌ ΔT = T f − T i ‌
the‌‌rocks‌‌is‌‌0.82‌‌J/g‌‌K.‌ ‌ ‌
‌ S:‌ H‌2‌O‌‌=‌2‌ .02‌+‌ ‌1‌ 6‌‌ ‌=‌‌18.02‌‌g /mol‌ ‌
G:‌ m‌‌=‌‌50.0‌‌kg‌ R:‌‌q‌ ‌ 2.75 mol · 18.02 g
1 mol = 49.555 g ‌
c‌‌=‌‌0.82‌‌J/g‌‌°C‌ ‌ ‌
ΔT‌‌=‌‌12.0‌‌°C‌ ‌ q = (49.555 g) (4.18 J /g °C) (98 °C − 22.0 °C) ‌
‌ q = (207.1399 J /°C) (76 °C) ‌ ‌
E:‌ q = mcΔT ‌ ‌
‌ q‌‌=‌1
‌ 5‌‌700‌‌J‌ ‌
1 000 g
S:‌ 50.0 kg · 1 kg = 50 000 g ‌ ‌
q = (50 000 g)(0.82 J /g°C)(12.0 °C) ‌ 4. A‌‌27.7‌‌g‌‌sample‌‌of‌‌ethylene‌‌glycol,‌‌a‌‌car‌‌radiator‌‌
‌ coolant,‌‌loses‌‌688‌‌J‌‌of‌‌heat.‌‌What‌‌was‌‌the‌‌initial‌‌
q‌‌=‌4
‌ 90‌‌000‌‌J‌ ‌ temperature‌‌of‌‌the‌‌ethylene‌‌glycol‌‌if‌‌the‌‌final‌‌
‌ temperature‌‌is‌‌32.5‌‌°C‌‌(c‌‌of‌‌ethylene‌‌glycol‌‌is‌‌2.42‌‌J/g‌‌
2. What‌‌temperature‌‌change‌‌would‌‌these‌‌rocks‌‌ K)?‌ ‌
undergo‌‌if‌‌they‌‌absorbed‌‌450‌‌kJ‌‌of‌‌heat?‌ ‌ ‌
‌ G:‌ m‌‌=‌‌27.7‌‌g‌ R:‌‌Ti‌‌ ‌
G:‌ m‌‌=‌‌50.0‌‌kg‌ R:‌‌ΔT‌ ‌ c‌‌=‌‌2.42‌‌J/g‌‌°C‌ ‌
c‌‌=‌‌0.82‌‌J/g‌‌°C‌ ‌ T‌f‌‌ ‌=‌‌32.‌‌5‌‌°C‌ ‌
q‌‌=‌‌450‌‌kJ‌ ‌ q‌‌=‌‌-‌‌688‌‌J‌ ‌
‌ ‌
q
E:‌ q = mcΔT → ΔT = mc ‌ E:‌ q = mcΔT → q = mc (T f − T i ) ‌
‌ ‌
50.0 kg · 1 1000 − 688 J = (27.7 g) (2.42 J /g °C) (32.5 °C − T i ) ‌
g
S:‌ kg = 50 000 g ‌ S:‌
450 kJ · 1 1000 J ‌ − 688 J = (67.034 J °C) (32.5 °C − T i ) ‌
kJ = 450 000 J (67.034 J °C) (32.5 °C −T i )
− 688 J ‌
‌ 67.034 J °C = 67.034 J °C

ΔT = 450 000 J
(50 000 g)(0.82 J /g °C) = 450 000 °C
41 000 ‌‌ − 10.26345 °C = 32.5 °C − T i ‌
‌ T i = 32.5 °C − 10.26345 °C ‌ ‌
ΔT‌‌=‌1
‌ 1‌‌°C‌ ‌ ‌
‌ T‌i‌‌ ‌=‌4
‌ 2.8‌‌°C‌ ‌‌
‌
‌