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WorksheetsHess's Law and Bomb Calorimetry
Total questions: 15
Worksheet time: 4hrs 45mins
Using the equations below:
C(s) + O2(g) → CO2(g) ∆H = –390 kJ
Mn(s) + O2(g) → MnO2(s) ∆H = –520 kJ
what is ∆H (in kJ) for the following reaction?
MnO2(s) + C(s) → Mn(s) + CO2(g)
910
130
-130
-910
Using the equations below
Cu(s) + 1/2O2(g) → CuO(s) ∆H = –156 kJ
2Cu(s) + O2(g) → Cu2O(s) ∆H = –170 kJ
what is the value of ∆H (in kJ) for the following reaction?
2CuO(s) → Cu2O(s) + 1/2O2(g)
142
15
-15
-142
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 following equations show the oxidation of carbon and carbon monoxide to carbon dioxide.
C(s) +O2(g) → CO2(g) ΔH = –x kJ mol–1
CO(g) + O2(g) → CO2(g) ΔH = –y kJ mol–1
What is the enthalpy change, in kJ mol–1, for the oxidation of carbon to carbon monoxide?
C(s) + O2(g) → CO(g)
x + y
-x - y
y - x
x - y
Which of the following statements are true for the reaction:
SO2(g) + 1/2O2(g) ↔ SO3(g)
ΔH = –92 kJ mol-1
Where ↔ indicates that the reaction can proceed in the forward and the reverse direction.
The forward and reverse reaction both produce 92 kJ of energy.
Oxidizing 2 moles of SO2 would produce twice as much energy.
The reverse reaction has an enthalpy of +92 kJ mol-1.
Collecting the SO3 produced in the liquid state would not change the measured enthalpy.
The standard enthalpy change of formation values of 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
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)
If N2 (g) + 2O2 (g) ⟶ 2NO2(g) has a ΔHrxn = 68, then what is the ΔHrxn if you reverse the reaction?
86 kJ
- 86 kJ
68 kJ
-68 kJ
How much energy is required to turn 1 mole of N2O4(g) into 2 moles N & 4 moles O(g)?
1933 kJ
-1875 kJ
- 1933 kJ
1875 kJ
-824.2
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
4a + 5b - c
2a + 10b - c
2a + 5b + c
Q= m c ∆T
The units for specific heat are:
-2745 kJ/mol
