WorksheetsEnthalpy of Reaction with Molar Values
Total questions: 11
Worksheet time: 12mins
How do you calculate Enthalpy of a reaction?
ΔH = ΔHproducts - ΔHreactants
ΔT = q / mC
ΔG = ΔH -TΔS
E = mc2
Define standard enthalpy of combustion.
Heat released when one mole of substance is burnt completely in excess oxygen.
Heat absorbed when one mole of substance is burnt completely in excess oxygen under standard state.
Heat released when one mole of substance is burnt completely in excess oxygen under standard state.
Heat change when one mole of substance is burnt partially in excess oxygen under standard state.
Which of the following has a ΔHfo value of 0?
Br2(g)
N(g)
CO(g)
Ne(g)
Which of the equation below refers to the standard enthalpy of formation, ΔHfo?
Na(g) ---> Na+(g) + e- ΔH = -364 kJmol-1
C2H5OH(l) + 3O2(g) ---> 2CO2(g) + 3H2O (l) ΔH = - 1286 kJmol-1
2C(s) + 2H2(g) ---> C2H4 (g) ΔH = - 52.3 kJmol-1
Na+(g) ---> Na+(aq) ΔH = - 364 kJmol-1
When 1.0 mole of ZnO(s) decomposes,
ZnO(s) ---> Zn(s) + 1/2 O2(g) , enthalpy change is +348 kJ/mol.
What does this tell you about the formation of ZnO (s)?
the formation of ZnO (s) is endothermic
the formation of ZnO (s) is exothermic
the formation of ZnO (s) does not require energy
the formation of ZnO (s) absorbs heat.
Ca(OH)2 -> CaO + H2O
ΔHf in kJ/mol:
Ca(OH)2 -983.2
CaO -634.9
H2O -285.5
The enthalpy change for the reaction
C(s, graphite) + 1⁄2O2(g) --> CO(g) cannot be measured directly since some carbon dioxide is always formed in the reaction.
It can be calculated using Hess’s Law and the enthalpy changes of combustion of graphite and of carbon monoxide.
C(s, graphite) + O2(g) --> CO2 ΔH=-394 kJmol–1
CO(g) + 1⁄2O2(g) --> CO2 ΔH=-283 kJmol–1
The enthalpy change for the reaction of graphite with oxygen to give carbon monoxide is
-677 kJmol–1
-111 kJmol–1
+111 kJmol–1
+677 kJmol–1
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
