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Summary AS level chemistry - CH2 - Unit 2 - WJEC (Wales) specification

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AS level chemistry - CH2 - Unit 2 - WJEC (Wales) specification: These notes are based on the WJEC curriculum but can be used for other specifications. Very in-depth in all topics (8 units) describing basic chemistry.

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‭Thermochemistry: Unit 2.1‬
‭Energetics‬
‭Enthalpy‬ ‭ he total energy content of a system at a constant pressure (KJmol‬‭-1‬‭)‬
T
‭Energy required to break 1 mole of bonds‬

‭Enthalpy change‬ ‭The change in energy (ΔH)‬

‭Exothermic‬ ‭ reaction where surroundings increase in temperature (the system has lost energy)‬
A
‭Negative enthalpy change (-ΔH)‬

‭Endothermic‬ ‭ reaction where surroundings decrease in temperature (the system has gained energy)‬
A
‭Positive enthalpy change (+ΔH)‬
‭Standard conditions‬
‭●‬ ‭Fixed conditions for comparing enthalpy change‬
‭○‬ ‭Enthalpy changes based on the conditions‬
‭●‬ ‭Elements that meet the standard conditions have 0 enthalpy‬
‭1.‬ ‭Temperature of 298K (25‬‭o‬‭)‬
‭2.‬ ‭Pressure of 1atm (101000Pa)‬
‭3.‬ ‭All substances are in their standard state‬
‭Standard enthalpy change‬ ‭ tandard molar enthalpy change where all substances meet the standard‬
S
‭(ΔH‬‭θ‬‭)‬ ‭conditions‬

‭Standard molar enthalpy‬ ‭ he enthalpy change when one mole of substance is formed from its‬
T
‭θ‬
‭change of formation (Δ‬‭f‭H
‬ ‬ ‭)‬ ‭constituent elements in their standard states under standard conditions‬

‭Standard molar enthalpy‬ ‭ he enthalpy change when one mole of substance in its standard state is‬
T
‭θ‬
‭change of combustion (Δ‬‭c‭H
‬ ‬ ‭)‬ ‭completely combusted in oxygen under standard conditions‬
‭Carbon dioxide and water are the only products‬
‭C‬‭2‭H ‬ ‬‭OH + 3O‬‭2‬ ‭→ 2CO‬‭2‬ ‭+ 3H‬‭2‬‭O‬
‬ ‭5


‭Standard enthalpy change of‬ T‭ he enthalpy change in a reaction between the number of moles of‬
‭θ‬
‭reaction (Δ‬‭r‭H
‬ ‬ )
‭ ‬ ‭reactants shown in the equation for the reaction under standard conditions‬
‭Calculated from standard enthalpy changes of formation of all reactants‬
‭and products‬
θ
‭-‬ ∆‭𝑟‭𝐻

‬ = Σ∆‭𝑓‬‭𝐻‬(‭𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠‬) − Σ∆‭𝑓‭𝐻 ‬
‬ ‭𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠‬

‭Hess’s law‬
‭●‬ ‭The total energy change that occurs when a system changes from one state to‬
‭another is independent of the route taken to bring about the change‬
‭●‬ ‭An interpretation of the law of conservation of energy‬




‭35‬

,‭Average bond enthalpy‬
‭Bond breaking‬ ‭An‬‭exothermic‬‭process - energy is absorbed‬

‭Bond formation‬ ‭An‬‭endothermic‬‭process - energy is evolved‬
‭●‬ ‭Enthalpy change for a particular bond depends on the structure of the molecule‬
‭○‬ ‭Therefore the values for bond enthalpy are usually an average‬
‭θ‬
‭Δ‭f‬‬‭H‬ ‭as an indication stability‬
‭θ‬
‭●‬ ‭The more negative the Δ‬‭f‭H
‬ ‬ ‭(exothermic) the easier‬‭the compound is to form‬
‭○‬ ‭More stable‬
‭Calorimetry‬
‭●‬ ‭Used to measure enthalpy changes‬
‭Temperature‬ ‭ he average kinetic energy of the particles in a system‬
T
‭Independant‬‭to the number present‬
‭Measured by a thermometer‬

‭Heat‬ ‭ he total energy in a given amount of substance‬
T
‭Dependant‬‭on how much of the substance is present‬
‭Measured using calorimetry‬

‭Specific heat capacity (c)‬ ‭The heat needed to raise the temperature of 1g of substance by 1K‬
‭●‬ ‭Measures the heat transferred to surrounding (not the heat of the system directly)‬
‭○‬ ‭We assume all heat is exchanged with the solution only‬




‭36‬

, ‭Δ‭r‬‬‭H between magnesium and sulphuric acid‬
‭Procedure‬ 1
‭ . Accurately weigh out approximately 0.11g of magnesium powder‬
‭2. Measure 50cm‬‭3‬ ‭of sulphuric acid using a burette‬‭and pour into a polystyrene cup‬
‭-‬ ‭Acid is in‬‭excess‬‭but the volume must be accurate‬‭to know the mass heated‬
‭3. Stir continuously using the thermometer and measure temperature every 30 seconds‬
‭-‬ ‭Must be at least 3 concordant readings‬
‭4. Add magnesium after 3.5 minutes‬
‭5. Stir continuously and measure temperature every 30 seconds until minute 7‬
‭6. Record results‬

‭Results‬ ‭ ass of acid = 50g (50cm‬‭3‬ ‭* 1gcm‬‭-3‬‭)‬
M
‭Specific heat capacity = 4.18‬
‭Temperature change = 9.7C‬‭o‬
‭Moles of magnesium = 4.53*10‬‭-3‬ ‭(0.11/24.3)‬
−(‭50‬×‭4.‬‭18‬×‭9‬.‭7‬)
∆‭𝐻‬ = −‭3‬
‭4‬.‭53‬×‭10‬




‭37‬

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