Science 10
Brakke
ECA – Chem – Topic 01
Worksheet Energy Diagrams – Answer Key
Name _________KEY_______________________
Directions: We are going to write what are called thermochemical equations. A thermochemical equation
includes the STATE of each substance, and how much energy is given off (in which case it goes on the product side) OR
taken in (in which case it goes on the reactants side).
These will all be word equations. You must translate the words into symbols and put the states on them-------(s) for solid, (l) for
liquid, (aq) for anything dissolved in water or referred to as a solution, and (g) for gas. Also, provide an energy diagram for each in the
table provided.
1.
When one mole of liquid water is produced from its elements 246 kJ of energy are given off. Write a balanced
thermochemical equation for this reaction and give the associated molar heat production as
well as the enthalpy for the specific reaction!
1.
2H2 (g) + O2 (g)
Equation: 2H2 (g) + O2 (g)  2H2O (l)
ΔHrxn =
ΔHH2O=
-492kJ
-246
kJ/mol
H2O (l)
2.
When phosphorus (P4) reacts with chlorine gas to produce phosphorus pentachloride 299 kJ
of energy are given off for each mole of phosphorus pentachloride that is produced. Write a
balanced thermochemical reaction for this.
2.
P4 (s) + 10 Cl2 (g)
Equation: P4 (s) + 10 Cl2 (g)  4 PCl5 (s)
ΔHrxn =
3.
-1196
kJ
ΔHPCl5=
-299kJ/mol
4 PCl5 (s)
When ammonium nitrate is dissolved in water 86 kJ of energy are absorbed or taken in by the
water for each mole of ammonium nitrate used. Write a balanced thermochemical reaction
for this process.
3.
NH4NO3 (l)
Equation: NH4NO3 (l)  NH4+ (aq) + NO3- (aq)
ΔHNH4NO3=
-86
kJ/mol
When methane, CH4, is burned in air 890 kJ of energy are given off per mole of methane.
The products of the reaction are gaseous carbon dioxide and liquid water. Write a balanced
thermochemical reaction for this process.
ΔHrxn =
4.
-86kJ
NH4+ (aq) + NO3- (aq)
4.
CH4 (g) + 2O2 (g)
Equation: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
ΔHrxn =
-890
kJ
ΔHCH4 =-890
kJ/mol
CO2 (g) + 2H2O (l)
5.
When one mole of sodium thiosulfate is dissolved in water 33.5 kJ of energy are absorbed by
the water. Write a balanced thermochemical reaction for this process.
Equation: Na2S2O3 (s)  2Na+ (aq) + S2O32- (aq)
ΔHrxn =-33.5
kJ
ΔHNa2S2O3=
5.
Na2S2O3 (s)
-33.5kJ/mol
2Na+ (aq) + S2O32- (aq)
Science 10
Brakke
ECA – Chem – Topic 01
DO NOT TURN OVER UNDER ANY CIRCUMSTANCES: KEEP THIS PAPER FLAT ON YOUR DESK (OR YOU GET A
ZERO FOR THE POP-QUIZ)
1. When 8.50 grams of sodium hydroxide dissolves in 250.0 grams of water in a coffee-cup calorimeter, the temperature rises
from 20.1oc to 29.5oc. Assume all heat absorbed in the solution process comes from the water.
a. Write a balanced equation for the solution process: (Remember, what does a salt do in water?)
NaOH (s)  Na+ (aq) + OH- (aq)
a. What is q of the water? Cannot find q of NaOH directly so must find it for H2O and use q = -q
qH2O = mcΔT
qH2O = (250g)( 4.184 J/goc)(9.4oc) = +9832 J
mH2O = 250 g
cH2O = 4.184 J/goc
ΔTH2O = 29.5-20.1 = 9.4oc
b. What is q when sodium hydroxide dissolves? Explain how you get this answer!
qNaOH = -(qH2O)
qNaOH = -(9842J)
qNaOH = -9832 J OR -9.832 kJ
c. Is the process endothermic or exothermic? Explain
q for the salt is negative (-) and therefore energy is released to the surroundings (water) making the heat change for
AgNO3 exothermic
d. What is the molar heat of solution when 8.50 grams ofsodium hydroxide dissolves?
−9832 𝐽
40.00𝑔 𝑁𝑎𝑂𝐻
𝐽
𝑘𝐽
𝑞𝑁𝑎𝑂𝐻 =
𝑥
= −46,268
= −46.27
= ΔH
8.50 𝑔 𝐴𝑔𝑁𝑂3
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
a. Draw an energy diagram for this reaction.
NaOH (s)
Enthalpy
ΔH = −46.27
Enthalpy
𝑘𝐽
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
Na+ (aq) + OH- (aq)
+
-
Na (aq) + OH (aq)
e. Level 5: Calculate the energy change (∆H)when 20molesof sodium hydroxide are dissolved in sufficient water:
20 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 𝑥
−46.27 𝑘𝐽
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= ΔH = -925.4 kJ
f. Level 6: Calculate the energy change (∆H) when 9.58 grams of sodium hydroxide are dissolved in sufficient water:
9.58 𝑔 𝑁𝑎𝑂𝐻 𝑥
1𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
40.00 𝑔 𝑁𝑎𝑂𝐻
𝑥
−46.27 𝑘𝐽
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= ΔH = - 11.08 kJ
f. Level 7: How many grams of sodium hydroxide were dissolved if the energy change (∆H) was 5.4 kilojoules?
5.4 𝑘𝐽 𝑥 |
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
40.00 𝑔 𝑁𝑎𝑂𝐻
|𝑥
= 4.67 𝑔 𝑁𝑎𝑂𝐻
−46.27 𝑘𝐽
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻