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5.1 Exothermic and endothermic reactions If a reaction produces heat (increases the temperature of the surroundings) it is exothermic. If a reaction produces heat (increases the temperature of the surroundings) it is exothermic. If the temperature of the reaction mixture decreases (ie heat is absorbed) then the reaction is endothermic. If the temperature of the reaction mixture decreases (ie heat is absorbed) then the reaction is endothermic. Exothermic -> a reaction which produces heat (ΔH has a negative value by convention, -ve) Exothermic -> a reaction which produces heat (ΔH has a negative value by convention, -ve) Endothermic -> a reaction which absorbs heat (ΔH has a positive value by convention, +ve) Endothermic -> a reaction which absorbs heat (ΔH has a positive value by convention, +ve)

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Enthalpy of reaction, ΔH: The change in internal (chemical) energy (H) in a reaction The most stable state is where all energy has been released. When going to a more stable state, energy will be released, and when going to a less stable state, energy will be gained (from the surroundings). The most stable state is where all energy has been released. When going to a more stable state, energy will be released, and when going to a less stable state, energy will be gained (from the surroundings). On an enthalpy level diagram, higher positions will be less stable (with more internal energy) therefore, if the product is lower, heat is released (more stable, ΔH is -ve) but if it is higher, heat is gained (less stable, ΔH is +ve). On an enthalpy level diagram, higher positions will be less stable (with more internal energy) therefore, if the product is lower, heat is released (more stable, ΔH is -ve) but if it is higher, heat is gained (less stable, ΔH is +ve).

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Formation of bonds causes an energy release (exothermic). Formation of bonds causes an energy release (exothermic). Breaking of bonds requires energy (endothermic). Breaking of bonds requires energy (endothermic).

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Calculation of enthalpy changes Change in energy = mass x specific heat capacity x change in temperature Change in energy = mass x specific heat capacity x change in temperature E = m x c x ΔT E = m x c x ΔT where m is the mass of water present (kilograms), and c = 4.18 kJ Kg-1 K- 1.

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Hess' Law Hess' Law states that the total enthalpy change between given reactants and products is that same regardless of any intermediate steps (or the reaction pathway). Hess' Law states that the total enthalpy change between given reactants and products is that same regardless of any intermediate steps (or the reaction pathway).

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Example: Calculate the enthalpy of formation of methane. Example 1: Data: ΔH combustion (carbon) = -394kj ΔH combustion (hydrogen) =-242kj ΔHcombustion (methane) =-891kj

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If ΔHf methane is represented by the equation: C(s) + H2(g) → CH4 (g) C(s) + H2(g) → CH4 (g) This equation can be constructed using the equations for combustion of the reactants (carbon and hydrogen) and the product (methane) C(s) + O2(g) → CO2(g) -394k …..enthalpy 1 C(s) + O2(g) → CO2(g) -394k …..enthalpy 1 H2(g) + ½ O2(g) → H2O(l) -242kj …..enthalpy 2 multiply this equation by 2 to get multiply this equation by 2 to get 2H2(g) + O2(g) → 2H2O(l) -484kj …….enthalpy 3 Add the first and third equation together to get: Add the first and third equation together to get: C(s) +2O2(g) +2H2(g) → CO2(g) + 2H2O(l) -878kj enthalpy (1 + 3) = 4 enthalpy (1 + 3) = 4 Now take away the eq for the combustion of methane Now take away the eq for the combustion of methane CH4 (g) + 2O2(g) → CO2(g) + 2H2O(l) -891kj enthalpy 5 CH4 (g) + 2O2(g) → CO2(g) + 2H2O(l) -891kj enthalpy 5 And after rearrangement ( take the CH4 to the right hand side) the result is the equation for the formation of methane And after rearrangement ( take the CH4 to the right hand side) the result is the equation for the formation of methane C(s) + H2(g) → CH4 (g) +13kj enthalpy ( 4 - 5) C(s) + H2(g) → CH4 (g) +13kj enthalpy ( 4 - 5)

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Example 1 Monoclinic sulphur is formed in volcanic regions by reaction between sulphur dioxide and hydrogen sulphide according to the equation: SO2 + 2H2S --> 2H2O + 3S SO2 + 2H2S --> 2H2O + 3S Draw an enthalpy diagram or cycle and calculate the standard enthalpy change for this reaction. Here are some values. Here are some values. Standard enthalpy of formation (all in kJ) Standard enthalpy of formation (all in kJ) H 2 O(l) -286 H 2 O(l) -286 H 2 S(g) - 20.2 H 2 S(g) - 20.2 Standard enthalpy of combustion of S (monoclinic) = -297.2kJ Standard enthalpy of combustion of S (monoclinic) = -297.2kJ equation 1: H 2 + ½ O 2 --> H 2 O -286 kJ equation 1: H 2 + ½ O 2 --> H 2 O -286 kJ equation 2: H 2 + S --> H2S -20.2kJ equation 2: H 2 + S --> H2S -20.2kJ equation 3: S + O 2 --> SO2 -297.2 kJ equation 3: S + O 2 --> SO2 -297.2 kJ multiply E2 by 2 … (equation 4) H 2 + 2S --> 2H 2 S -40.4 kJ multiply E2 by 2 … (equation 4) H 2 + 2S --> 2H 2 S -40.4 kJ add equation 3 & 4 …. ….. 2H 2 + 3S + O 2 --> SO 2 + 2H 2 S -337.6 kJ add equation 3 & 4 …. ….. 2H 2 + 3S + O 2 --> SO 2 + 2H 2 S -337.6 kJ multiply E1 by 3 … (equation 5) 2H 2 + O 2 --> 2H 2 O -572 kJ multiply E1 by 3 … (equation 5) 2H 2 + O 2 --> 2H 2 O -572 kJ subtract 4 from 5 …… -3S ---> 2H2O - SO2 - 2H2S -234.4 kJ subtract 4 from 5 …… -3S ---> 2H2O - SO2 - 2H2S -234.4 kJ rearrange SO 2 + 2H 2 S --> 2H 2 O + 3S -234.4 kJ

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Born Haber Cycles Born Haber cycles are the application of Hess' law to ionic systems. An ionic solid consists of a giant structure of ions held together in a giant lattice. Born Haber cycles are the application of Hess' law to ionic systems. An ionic solid consists of a giant structure of ions held together in a giant lattice. Application of Hess law tells us that the enthalpy of formation of an ionic crystal is equal to the sum of the energies of formation of the ions plus the enthalpy of the lattice. Application of Hess law tells us that the enthalpy of formation of an ionic crystal is equal to the sum of the energies of formation of the ions plus the enthalpy of the lattice. it is a several step process that is best represented by a diagram showing the individual steps as endothermic upwards and exothermic downwards. it is a several step process that is best represented by a diagram showing the individual steps as endothermic upwards and exothermic downwards.

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