WorksheetsThermo REVIEW
Total questions: 25
Worksheet time: 1hrs 18mins
How do you calculate Enthalpy of a reaction?
ΔH = ΔHproducts - ΔHreactants
ΔT = q / mC
ΔG = ΔH -TΔS
E = mc2
Which statement is true regarding ENDOTHERMIC reactions?
Temperature change is positive and enthalpy change is positive
Temperature change is positive and enthalpy change is negative
Temperature change is negative and enthalpy change is positive
Temperature change is negative and enthalpy change is negative
The enthalpy change is a negative (-) value, what side of the equation will the energy value be on.
Reactants
Products
endothermic reaction
exothermic reaction
In the Image, will the heat (+177.8 KJ) be a product or a reactant?
Product
Reactant
Define the term Exothermic by picking the correct statements. (Choose 3)
Products have less energy than the reactants
ΔH is negative
ΔH is positive
Energy is given out to the surroundings (temperature goes up).
products have more energy than the reactants
Define Endothermic by picking the correct statements. (Choose 3)
ΔH is positive
Energy is taken in from the surroundings (temperature goes down).
Energy is given out to the surroundings (temperature goes up).
· Products have less energy than the reactants
products have more energy than the reactants
Is this equation endo- or exo-thermic?
PCl3 (s) + Cl2 (g) --> PCl5 (s) + energy
endothermic
exothermic
How much heat in KJ does an aluminum block absorb if 100.0 grams are heated from 25.0oC to 50.0oC? the specific heat of aluminum is 0.900J/goC
4.5 KJ
-4.5 KJ
2.25 KJ
-2.25 KJ
A sample of iron receives 50.J of heat energy that raises the temperature of the iron by 25.0°C. If iron has a specific heat of 0.10 J/g°C, what is the mass of the iron sample?
25g
30g
20g
50g
Based on the information in the table above, which of the following expressions gives the approximate ΔH° for the reaction represented by the following balanced chemical equation?
Fe2O3(s)+3 CO(g)→2 Fe(s)+3 CO2(g)
ΔH°rxn=[(0kJ/mol)+(−394kJ/mol)]−[(−826kJ/mol)+(−111kJ/mol)]
ΔH°rxn=[2(0kJ/mol)+ 3(−394kJ/mol)]−[(−826kJ/mol)+ 3(−111kJ/mol)]
ΔH°rxn=[(−826kJ/mol)+ 3(−111kJ/mol)]−[2(0kJ/mol)+ 3(−394kJ/mol)]
ΔH°rxn=[(−826kJ/mol)+(−111kJ/mol)]−[(0kJ/mol)+(−394kJ/mol)]
Calculate the enthalpy of reaction (ΔH).
4CO2 (g) + 6H2O (g) → 2C2H6 (g) + 7O2 (g)
ΔHfo [CO2(g)] = -393.5 kJ
ΔHfo [H2O(g)] = -241.8 kJ
ΔHfo [C2H6(g)] = -84.7 kJ
550.647 KJ/mol
2855.584 KJ/mol
-550.647 KJ/mol
-2855.584 KJ/mol
Consider the following equations.
Mg(s) + O2(g) → MgO(s) ∆H = –602 kJ
H2(g) + O2(g) → H2O(g) ∆H = –242 kJ
What is the ∆H value (in kJ) for the following reaction?
MgO(s) + H2(g) → Mg(s) + H2O(g)
-844
-360
+360
+844
The enthalpies of combustion of C(s), H2(g) and C4H9OH(l) (in kJmol-1) are as follows
C(s) + O2(g) -> CO2(g) ∆H=a
H2(g) + ½O2(g) -> H2O(l) ∆H=b
C4H9OH(l) + 6O2(g) -> 4CO2(g) + 5H2O(l) ∆H=c
What is the enthalpy change for the reaction shown below?
4C(g) + 5H2(l) + ½O2(g) -> C4H9OH(l)
c – 4a – 5b
2a + 10b - c
4a + 5b - c
2a + 5b + c
4 NH3 (g) + 5 O2 (g) ⟶ 4 NO (g) + 6 H2O (g)
Using the following information, calculate ΔH for the reaction shown above:
N2 (g) + O2 (g) ⟶ 2 NO (g) ΔH = -180.5 kJ
N2 (g) + 3 H2 (g) ⟶ 2 NH3 (g) ΔH = -91.8 kJ
2 H2 (g) + O2 (g) ⟶ 2 H2O (g) ΔH = -483.6 kJ
256.0 kJ
-1628.2 kJ
-6387 kJ
None of these
The enthalpy of reactions for the synthesis two oxides of phosphorus are:
P4(s) + 3O2(g) → P4O6(s) ΔHf = –1600 kJ mol–1
P4(s) + 5O2(g) → P4O10(s) ΔHf = –3000 kJ mol–1
What is the enthalpy change, in kJ mol–1, for the reaction below?
P4O6(s) + 2O2(g) → P4O10(s)
+4600
+1400
–1400
–4600
A piece of metal with a mass of 32.8 g is heated to 100.5°C and dropped into 138.2 g of water at 20.0°C. The final temperature of the system is 30.2°C. What is the specific heat capacity of the metal? (show your work)
2.56 J/g°C
0.391 J/g°C
5.29 J/g°C
3.50 J/g°C
What does ΔT represent?
tf −ti
ti −tf
t - celsius
celsius - t
How much energy must be used to produce 4.75 mol of gaseous water?
H2O (l) + 44.0 kJ --> H2O (g)
(Remember, coefficient of 1 = 1 mole)
207 kJ
9.36 kJ
206.8 kJ
9.362 kJ
300 mL of water at 95 ºC are added to 700 mL of water at 62.8 ºC. What is the final temperature of the full liter of water?
50 ºC
85.6 ºC
72.5 ºC
94.2 ºC
P4 + 6Cl2 --> 4PCl3 + 2439 kJ
