|Matters and Properties of Matters|
|Atomic Structure with Examples|
|The Mole Concept with Examples|
|Gases with Examples|
|Chemical Reactions with Examples|
|Nuclear Chemistry (Radioactivity)|
|Acids and Bases|
|Rates of Reactions (Chemical Kinetics)|
|Exams and Problem Solutions|
|--Matters and Properties of Matters Exams and Problem Solutions|
|--Atomic Structure Exams and Problem Solutions|
|--Periodic Table Exams and Problem Solutions|
|--The Mole Concept Exams and Problem Solutions|
|--Gases Exams and Problem Solutions|
|--Chemical Reactions Exams and Problem Solutions|
|--Nuclear Chemistry (Radioactivity) Exams and Problem Solutions|
|--Solutions Exams and Problem Solutions|
|--Acids and Bases Exams and Problem Solutions|
|--Thermochemistry Exams and Problem Solutions|
|--Thermochemistry Exam1 and Problem Solutions|
|--Thermochemistry Exam2 and Problem Solutions|
|--Rates of Reaction Exams and Problem Solutions|
|--Chemical Equilibrium Exams and Problem Solutions|
|--Chemical Bonds Exams and Problem Solutions|
Thermochemistry Exam1 and Problem Solutions
1. Which ones of the following reactions are endothermic in other words ∆H is positive?
I. H2O(l) + 10,5kcal → H2O(g) ∆H1
II. 2NH3 +22kcal →N2 + 3H2 ∆H2
III. Na + Energy → Na+1 + e- ∆H3
When matters change state from liquid to gas, they absorb energy. I is endothermic reaction. ∆H1 is positive.
In decomposition reactions energy (heat) is absorbed. III is endothermic reaction. ∆H2 is positive.
To remove one electron from atom we should give energy, so III is endothermic reaction and ∆H3 is positive.
2. Given table shows standard molar enthalpy of formation of some matters.
Find enthalpy of C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) using data given in the table below.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
∆H=[3∆HCO2 + 4∆HH2O] - [1∆HC3H8 + 5∆HO2]
Since O2 is element, molar formation enthalpy of it is zero.
∆H=[3.(-94) + 4.(-60)] - [1.(-25) + 5.0]
∆H=-522 + 25
∆H=-497 kcal/mol (it is negative, in other words reaction is exothermic)
3. To calculate enthalpy of ; CO2(g) + H2(g) → CO(g) + H2O(g) which ones of the following must be known?
I. Molar formation enthalpy of H2O(g)
II. Molar formation enthalpies of CO(g) and CO2(g)
III. Enthalpy of reaction; H2(g) + 1/2O2(g) → H2O(g)
We find enthalpy of CO2(g) + H2(g) → CO(g) + H2O(g);
∆H=Σa∆H(F.(Products) - Σb∆H(F.(Reactants)
∆H=[∆HCO + ∆HH2O] - [∆HCO2 + ∆HH2]
Since H2 is element, molar formation enthalpy of it zero.
So, we must know I and II to find enthalpy of given reaction.
4. Find molar combustion enthalpy of C2H5OH using following molar enthalpies of matters;
∆H C2H5OH(l)= -67 kcal/mol
∆H CO2(g)= -94 kcal/mol
∆H H2O(l)= -68 kcal/mol
We should first write combustion reaction of C2H5OH;
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(s)
We use following formula to find unknown enthalpy;
∆HReaction=Σa∆H(Products) - Σb∆H(Reactants)
∆HCombustion=(2∆HCO2(g) + 3 ∆HH2O(l) ) - (∆HC2H5OH(l) + 3∆HO2)
∆HCombustion=[2.(-94) + 3.(-68)] - [-67]
∆HCombustion= -325 kcal/mol
5. There are 32 g S in 1000 g vitreous calorimeter having 1000 g water in it. If 32 g S is burned up in calorimeter, temperature rises from 20 0C to 90 0C. Find molar combustion enthalpy of S.
We find heat gained by glass and water during combustion by formula;
Qcalorimeter=70000 + 14000= 84000 cal
1 mol S is 32 g.
Molar combustion enthalpy of S is 84000 cal or 84 kcal.
Since it is combustion enthalpy;
∆HCombustionS= -84 kcal/mol