AQA AS Chemistry Unit 2 3.2.1 Energetics

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3.2.1 Energetics
Enthalpy Change (H)
Enthalpy change ( H) is the amount of heat energy released or absorbed when a chemical or physical change occurs
at constant pressure.
Standard enthalpy changes refer to standard conditions, 100kPa and 298K.
Enthalpy change can be positive or negative (endothermic and exothermic)
Endothermic is the gain of heat energy by a system (positive).
Exothermic is the loss of heat energy by a system (negative).
Standard Enthalpy of Combustion ( Hc ) is the enthalpy change, under standard conditions, when 1 mol of a
substance is burned completely in oxygen, with all reactants and products in their standard states.
Standard Enthalpy of Formation ( Hf ) is the enthalpy change, under standard conditions, when 1 mol of a compound
is formed from its elements, with all reactants and products in their standard states.
By definition for an element, the standard enthalpy of formation is zero.
Calorimetry
q = mcT
q is the heat energy (kJ).
m is the mass of the substance (g).
c is the specific heat capacity (kJK-1g-1).
T is the change in temperature (K).
E.g. In an insulated container, 50cm3 of 2.00mol dm-3 HCl at 293K were added to 50.0cm3 of 2.00mol dm-3 NaOH also
at 293K. After reaction, the temperature of the mixture rose to 307K. The specific heat capacity of water is 4.18 kJ K-1
kg-1.
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
Calculate the enthalpy change.
T = 14.0K
q=mc T
q=0.100 x 4.18 x 14.0
q=5.85kJ
Heat energy lost by the reaction = -5.85kJ
50
Moles of acid =VC = 2.00×1000 = 0.100 mol
heat lost by the reaction -1
Enthalpy change = moles of acid = -5.85
0.100 =- 58.5 kJ mol
Simple Applications of Hess's Law
The first law of thermodynamics states that energy can be neither created nor destroyed, but can be converted
from one form into another.
Hess's Law states that the enthalpy change of a reaction depends only on the initial and final states of the reaction
and is independent of the route by which the reaction occurs.

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E.g. Use the following data to calculate the enthalpy of formation, Hf, of ethanol, C2H5OH.
HcC2H5OH = -1367.3 kJ mol-1
HcC = -393.5 kJ mol-1
HcH2 = -285.8 kJ mol-1
2C(s) + 3H2(g) + ½O2(g) à C2H5OH(l)
2CO2(g) + 3H2O(l)
Clockwise = anticlockwise
Hf-1367.3 =(-393.5x2) + (-285.8x3)
Hf= 1367.3 - 857.4 - 787
Hf= -277.…read more

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