UI CHEM 001 Best Chemistry Thermochemistry Solutions Practice Exam

Best Chemistry Thermochemistry Solutions Practice Exam
University of Indonesia CHEM 001
20’s Thermochemistry Practice Exam Solutions


Thermochemistry

Exothermic and Endothermic Reactions

Reactant Product
HR Hp Enthalpy
Enthalpy Change: H = HP – HR
Exothermic Reaction: H = HP – HR < 0
Endothermic Reaction: H = HP – HR > 0




Thermochemistry Equation

H2(g) + ½ O2(g) H2O (l) H = -286 kJ
Especially in thermochemistry equations Coefficient = Mole

If the coefficient of a thermochemistry equation is multiplied by a certain
number, then the H is also multiplied by that number:
H2(g) + ½ O2(g) H2O (l) H = -286 kJ
2 H2(g) + O2(g) 2 H2O (l) H = -572 kJ

If the thermochemical equation is reversed, then H is given the opposite sign
H2(g) + ½ O2(g) H2O (l) H = -286 kJ
H2O (l) H2(g) + ½ O2(g) H = +286 kJ

,Molar Enthalpy

Standard Enthalpy of Formation (Hf°)(25°C, 1atm)
Hf° CO2(g) = -394 kJ/mole
C(s) + O2(g) CO2(g) H = -394 kJ/mole

Standard Enthalpy of Decomposition (Hd°)(25°C, 1atm) Hd° = - Hf°
Hd° CO2(g) = +394 kJ/mole
CO2(g) C(s) + O2(g) H = +394 kJ/mole

Standard Enthalpy of Combustion (Hc°)(25°C, 1atm) Generally exothermic
Hc° H2(g) = -286 kJ/mole
H2(g) + ½ O2(g) H2O (l) H = -286 kJ/mole
So, Hc° H2(g) = Hf° H2O(l)

Enthalpy of Neutralization (Hn°)
The enthalpy change for the reaction between an acid (H+) and a base (OH-)
forms 1 mole of water
HCN(aq) + NaOH(aq) NaCN(aq) + H2O(l) H = -12 kJ/mole

Enthalpy of Fusion (Hfus°)
H2O (s) H2O(l) H = +6 kJ/mole

Enthalpy of Solution (Hs°)
NaCl(s) Na+(aq) + Cl-(aq) H = +4 kJ/mole

, Determination of H by experiment (Calorimeter)

Q = amount of calories
Qsolution = mc T
m = the mass of the solution in the calorimeter
Qcalorimeter = C T
c = specific heat of solution in the calorimeter
Qreaction = Qsolution + Qcalorimeter C = heat capacity of the calorimeter
H = - Qreaction T = temperature change in the calorimeter
𝐻 H = reaction enthalpy change
H°=
𝑛 H°= standard enthalpy change for the reaction
n = moles of substance involved




The determination of H based on standard formation enthalpy
change data (Hf°)


aA + bB cC + dD H = ?

H° = (c. Hf°C + d. Hf°D) – (a. Hf°A + b. Hf°B)

H° = Hf° product - Hf° reactant
Hf° for free elements is zero