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Thermochemistry, Hess law PPTxx 100L.pptx

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1. Thermochemistry investigates heat flow into or out of reactions and the energy stored in chemical bonds. The heat absorbed or released by a reaction at constant temperature and pressure is equal to the change in enthalpy, ΔH. 2. Hess's law states that the enthalpy change of a reaction is constant regardless of the reaction path. The enthalpy change of a reaction can be calculated from the sum of enthalpy changes of each step. 3. Hess's law allows calculation of enthalpy changes that cannot be measured experimentally, such as heat of formation, through combination of other reaction enthalpies according to algebraic equations.

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PHYSICAL CHEMISTRY CHM 100L/CHEMISTRY DEPARTMENT DR. MRS BAMGBOSE 1
THERMOCHEMISTRY Thermochemistry is the branch of thermodynamics that investigates the heat flow into or out of a reaction system and deduces the energy stored in chemical bonds. As reactants are converted into products energy can either be taken up by the system or released to the surroundings. At constant temperature and constant volume, the heat that flows to the reaction system is equal to ∆𝑈. For a reaction that takes place at constant temperature and constant pressure, the heat that flows to the system is equal to ∆𝐻 for the reaction. The enthalpy of formation is defined as the heat flow into the system in a reaction between pure elements that leads to the formation of 1 mol of product. Because it is a state function, the reaction enthalpy can be written as the enthalpies of formation of the products minus those of the reactants. This property allows ∆𝐻 𝑎𝑛𝑑 ∆𝑈 for the reaction to be calculated for many reactions without carrying out an experiment.
Such that: ∆𝑈 = 𝑈2 − 𝑈1 or 𝑈𝑝𝑟𝑑𝑡−𝑈𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 ∆𝐻𝑚 ° = 𝐻𝑚𝑝𝑟𝑜𝑑𝑢𝑐𝑡 ° − 𝐻𝑚𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 ° For the reaction F𝑒3𝑜4 (𝑠) +4𝐻2(𝑔) →3 F𝑒(𝑠)+4𝐻2𝑂 ∆𝐻𝑚 ° =[𝐻° (3 F𝑒(𝑠)+𝐻° (4𝐻2𝑂4 𝑙)] −[𝐻° (𝐹𝑒3𝑜4 (𝑠)+𝐻° (4𝐻2(𝑔))]
HESS LAW Hess law or Hess’s law of constant heat summation states that at constant temperature heat energy changes (enthalpy) accompanying a chemical reaction will remain constant, irrespective of the way the reactants react to form products. This law is based on the state function character of enthalpy and the first law of thermodynamics. The energy (enthalpy) of a system (molecule) is a state function. So enthalpy of reactant and product molecules is a constant and does not change. According to the first law of thermodynamics the total energy of the substances before and after any (physical or chemical) charge should be equal. According to the law, the total energy of the reactant should be equal to the total energy of the product. Any difference in the energy between the reactants and products is also fixed at a particular enthalpy changes of all those chemical reactions.
By illustration:- The total enthalpy change (∆𝐻) of a reaction equals to the sum of the enthalpy changes occurring in each step of the reaction. i.e Figure 1(a). ∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = ∆𝐻1+∆𝐻2 Reactants Products Intermediates ∆ ∆ 2 ∆ 1
A B . D E Figure b. intermediates ∆𝐻5 intermediates ∆ ∆ 4 ∆ 1
Reactants can form product B by following three different steps C, D and E are intermediates in the other stepwise reactions. Hess law states that the enthalpy of the reaction ∆𝐻1 is the same irrespective of the path. So, the enthalpy of direct single step reaction and other paths given intermediates C, D and E should be the same. ∆𝐻1=∆𝐻2+∆𝐻3=∆𝐻4+∆𝐻5+∆𝐻 e.g C(s) + 𝑂2 𝑔 → C 𝑂 𝑔 , +26.0 kcal … (i) 𝐶𝑂 𝑔 + 𝑂 𝑔 → C𝑂2(g) , +68.3 kcal …(ii)
On adding the two reactions (i) + (ii) 𝐶 𝑠 +𝑂2 𝑔 → C𝑂2 𝑔 +94,3 kcals (26.0, 68.3 kcal) For Hess law ∆𝐻 = ∆𝐻1 + ∆𝐻2= (26+68.3 kCal) This means that carbon reacts with oxygen to form carbon dioxide releasing 94.3 kcals of heat in a single step. Carbon can also react in a two step process of forming an intermediate carbon monoxide, which again is converted to carbon dioxide. Net reaction enthalpy of both reactions is the same as that of single step formation. So, enthalpy of reaction does not change on the path followed by the reactants.
HESS LAW AND MULTI DIFFERENT REACTIONS Combustion of carbon, sulphur and carbon disulphide are exothermic with an enthalpy of -393.5kJ, -296.8kJ and -1075kJ. The reactants are: 𝐶 𝑠 + 𝑂2 𝑔 → C𝑂2 𝑔 +393.6 kJ …..(i) 𝑆 𝑠 + 𝑂2(𝑔)→ S𝑂2 𝑔 +296.8 kJ …..(ii) C𝑆2(𝑙)+ 3𝑂2 𝑠 → C𝑂2 𝑔 + 2S𝑂2 𝑔 +1075.0kJ …(iii) These reactions and enthalpy changes can be treated as algebraic equations to get the heat of formation of carbon disulphide even without experiments.
Equation (i) 𝐶 𝑠 +𝑂2 𝑠 → C𝑂2 𝑔 , +393.5 kJ Multiply eq (ii) by 2. Gives: 2𝑆2(𝑙) + 𝑂2 𝑔 → 2S𝑂2 𝑔 +593.6kJ Reverse equation (iii) to give C𝑂2 𝑔 + 𝑆𝑂2 𝑔 → C𝑆2(𝑙) + 3𝑂2 𝑔 -1075.0kJ Add the three reactions 𝐶 𝑠 + 𝑆 𝑠 → C𝑆2(𝑙) −87.9kJ Formation of carbon disulphide is an exothermic reaction.
APPLICATION OF HESS LAW OF HEAT SUMMATION 1. Enthalpy change in physical change Carbon and diamond are allotropes of carbon. But measuring the energy change in the conversion of graphite to diamond cannot be determined as the process cannot be carried out. The heat changes for this hypothetical physical change can be calculated using Hess law because: Hess law of heat summation is an efficient way to estimate heat changes that cannot be measured experimentally let’s consider the combination of graphite and diamond with oxygen with the heat of reaction as -393 and -395.4kJ respectively. C (graphite) + 𝑂2→ C𝑂2, ∆𝐻𝑔 = -393.4 kJ (i) C (diamond) + 𝑂2→ C𝑂2 , ∆𝐻𝑑 = -395.4kJ (ii)
Reverse equation (ii) as C𝑂2 → C (diamond) + 𝑂2, ∆𝐻𝑑= 395.4kJ (iii) Then add eqn (i) and (iii): C (graphite) + 𝑂2→ ∆𝐻𝑔𝑟 -393.4KJ C (graphite)→C (diamond) ∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = +2kJ Enthalpy change in the allotrope transition of graphite to diamond is endothermic of 2KJ.
2. Enthalpy change of chemical reaction The bond energy of hydrogen, iodine and hydrogen idodie are 218, 107KJ and 299KJ respectively. Question:- Estimate the enthalpy of hydrogen iodine formation. Is the reaction endothermic or exothermic? Consider the formation of hydrogen iodide from hydr ogen and iodide: 1 2 𝐻2(𝑔) + 1 2 𝐼2(𝑔) → H𝐼(𝑔) Enthalpy of formation of hydrogen, iodide is the heat change occurring when one atom of hydrogen and one atom of iodine react to form one mole of hydrogen iodide in standard conditions (as gas). To get one atom of hydrogen or iodine, the molecular bond has to be broken. Heat of formation = bond energy of HI - Bond dissociation of 𝐻𝑔 - Bond dissociation energy of 𝐼2 = 299-(218++107) = 299-325 = -26KJ As the heat of formation is negative, the reaction is exothermic.
3. Enthalpy of formation Carbon combines with hydrogen to form many hydrocarbons. Hence, the heat of formation of benzene cannot be determined experimentally but the heat change can be calculated by Hess law 6C + 3𝐻2 → 𝐶 𝐻 , ∆𝐻𝐶6𝐻6 =? Heat of formation of 𝐶𝑂2 and 𝐻2𝑂 are -393.5k and -285KJ respectively. Heat of combustion of benzene is -330KJ such that: C+𝑂2 → C𝑂2 , ∆𝐻1 = -393.5kJ …..(1a) H + 𝑂2 → 𝐻2𝑂 , ∆𝐻2 = -285.8kJ …..(2a) . 𝐶 𝐻 + 9𝑂2 → 6 C𝑂2+3𝐻2𝑂, ∆𝐻3= -3301 kJ (3a) Multiply reaction (i) by 6 6C+6𝑂2 →6C𝑂2, 6∆𝐻1 = -2361kJ (Ib) Multiply reaction (2a) by 3, will give 3𝐻2 + 3𝑂2 → 3𝐻2𝑂, 3∆𝐻2= -857.4kJ (2b)
6C 𝑂2 + 3𝐻2𝑂 → 6C 𝑂2 , 𝐶 𝐻 + 9 𝑂2, −∆𝐻3 = −∆𝐻3 = 3301kJ …(3b) Adding the three reactions give: 6C + 3𝐻2 → 𝐶 𝐻 , ∆𝐻 =82.6kJ The heat of formation of benzene =82.6kJ The heat of formation of benzene = +82.6kJ
kb Assignment I The heat of formation of carbon monozide and steam are 111 and 244KJ respectively. Calculate the heat of the reaction: 𝐻2𝑂 + C = CO + 𝐻2 2. Define heat of formation. 3. The reaction between gaseous hydrogen and gaseous Chlorine to form gaseous hydrogen chloride is given as: 𝐻2(g) + Cl(g) 2HCl(g) ΔH = -184.22 kJ . Calculate the heat of formation of Hydrogen Chloride and write the appropriate equation.

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Thermochemistry, Hess law PPTxx 100L.pptx

  • 1. PHYSICAL CHEMISTRY CHM 100L/CHEMISTRY DEPARTMENT DR. MRS BAMGBOSE 1
  • 2. THERMOCHEMISTRY Thermochemistry is the branch of thermodynamics that investigates the heat flow into or out of a reaction system and deduces the energy stored in chemical bonds. As reactants are converted into products energy can either be taken up by the system or released to the surroundings. At constant temperature and constant volume, the heat that flows to the reaction system is equal to ∆𝑈. For a reaction that takes place at constant temperature and constant pressure, the heat that flows to the system is equal to ∆𝐻 for the reaction. The enthalpy of formation is defined as the heat flow into the system in a reaction between pure elements that leads to the formation of 1 mol of product. Because it is a state function, the reaction enthalpy can be written as the enthalpies of formation of the products minus those of the reactants. This property allows ∆𝐻 𝑎𝑛𝑑 ∆𝑈 for the reaction to be calculated for many reactions without carrying out an experiment.
  • 3. Such that: ∆𝑈 = 𝑈2 − 𝑈1 or 𝑈𝑝𝑟𝑑𝑡−𝑈𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 ∆𝐻𝑚 ° = 𝐻𝑚𝑝𝑟𝑜𝑑𝑢𝑐𝑡 ° − 𝐻𝑚𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 ° For the reaction F𝑒3𝑜4 (𝑠) +4𝐻2(𝑔) →3 F𝑒(𝑠)+4𝐻2𝑂 ∆𝐻𝑚 ° =[𝐻° (3 F𝑒(𝑠)+𝐻° (4𝐻2𝑂4 𝑙)] −[𝐻° (𝐹𝑒3𝑜4 (𝑠)+𝐻° (4𝐻2(𝑔))]
  • 4. HESS LAW Hess law or Hess’s law of constant heat summation states that at constant temperature heat energy changes (enthalpy) accompanying a chemical reaction will remain constant, irrespective of the way the reactants react to form products. This law is based on the state function character of enthalpy and the first law of thermodynamics. The energy (enthalpy) of a system (molecule) is a state function. So enthalpy of reactant and product molecules is a constant and does not change. According to the first law of thermodynamics the total energy of the substances before and after any (physical or chemical) charge should be equal. According to the law, the total energy of the reactant should be equal to the total energy of the product. Any difference in the energy between the reactants and products is also fixed at a particular enthalpy changes of all those chemical reactions.
  • 5. By illustration:- The total enthalpy change (∆𝐻) of a reaction equals to the sum of the enthalpy changes occurring in each step of the reaction. i.e Figure 1(a). ∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = ∆𝐻1+∆𝐻2 Reactants Products Intermediates ∆ ∆ 2 ∆ 1
  • 6. A B . D E Figure b. intermediates ∆𝐻5 intermediates ∆ ∆ 4 ∆ 1
  • 7. Reactants can form product B by following three different steps C, D and E are intermediates in the other stepwise reactions. Hess law states that the enthalpy of the reaction ∆𝐻1 is the same irrespective of the path. So, the enthalpy of direct single step reaction and other paths given intermediates C, D and E should be the same. ∆𝐻1=∆𝐻2+∆𝐻3=∆𝐻4+∆𝐻5+∆𝐻 e.g C(s) + 𝑂2 𝑔 → C 𝑂 𝑔 , +26.0 kcal … (i) 𝐶𝑂 𝑔 + 𝑂 𝑔 → C𝑂2(g) , +68.3 kcal …(ii)
  • 8. On adding the two reactions (i) + (ii) 𝐶 𝑠 +𝑂2 𝑔 → C𝑂2 𝑔 +94,3 kcals (26.0, 68.3 kcal) For Hess law ∆𝐻 = ∆𝐻1 + ∆𝐻2= (26+68.3 kCal) This means that carbon reacts with oxygen to form carbon dioxide releasing 94.3 kcals of heat in a single step. Carbon can also react in a two step process of forming an intermediate carbon monoxide, which again is converted to carbon dioxide. Net reaction enthalpy of both reactions is the same as that of single step formation. So, enthalpy of reaction does not change on the path followed by the reactants.
  • 9. HESS LAW AND MULTI DIFFERENT REACTIONS Combustion of carbon, sulphur and carbon disulphide are exothermic with an enthalpy of -393.5kJ, -296.8kJ and -1075kJ. The reactants are: 𝐶 𝑠 + 𝑂2 𝑔 → C𝑂2 𝑔 +393.6 kJ …..(i) 𝑆 𝑠 + 𝑂2(𝑔)→ S𝑂2 𝑔 +296.8 kJ …..(ii) C𝑆2(𝑙)+ 3𝑂2 𝑠 → C𝑂2 𝑔 + 2S𝑂2 𝑔 +1075.0kJ …(iii) These reactions and enthalpy changes can be treated as algebraic equations to get the heat of formation of carbon disulphide even without experiments.
  • 10. Equation (i) 𝐶 𝑠 +𝑂2 𝑠 → C𝑂2 𝑔 , +393.5 kJ Multiply eq (ii) by 2. Gives: 2𝑆2(𝑙) + 𝑂2 𝑔 → 2S𝑂2 𝑔 +593.6kJ Reverse equation (iii) to give C𝑂2 𝑔 + 𝑆𝑂2 𝑔 → C𝑆2(𝑙) + 3𝑂2 𝑔 -1075.0kJ Add the three reactions 𝐶 𝑠 + 𝑆 𝑠 → C𝑆2(𝑙) −87.9kJ Formation of carbon disulphide is an exothermic reaction.
  • 11. APPLICATION OF HESS LAW OF HEAT SUMMATION 1. Enthalpy change in physical change Carbon and diamond are allotropes of carbon. But measuring the energy change in the conversion of graphite to diamond cannot be determined as the process cannot be carried out. The heat changes for this hypothetical physical change can be calculated using Hess law because: Hess law of heat summation is an efficient way to estimate heat changes that cannot be measured experimentally let’s consider the combination of graphite and diamond with oxygen with the heat of reaction as -393 and -395.4kJ respectively. C (graphite) + 𝑂2→ C𝑂2, ∆𝐻𝑔 = -393.4 kJ (i) C (diamond) + 𝑂2→ C𝑂2 , ∆𝐻𝑑 = -395.4kJ (ii)
  • 12. Reverse equation (ii) as C𝑂2 → C (diamond) + 𝑂2, ∆𝐻𝑑= 395.4kJ (iii) Then add eqn (i) and (iii): C (graphite) + 𝑂2→ ∆𝐻𝑔𝑟 -393.4KJ C (graphite)→C (diamond) ∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = +2kJ Enthalpy change in the allotrope transition of graphite to diamond is endothermic of 2KJ.
  • 13. 2. Enthalpy change of chemical reaction The bond energy of hydrogen, iodine and hydrogen idodie are 218, 107KJ and 299KJ respectively. Question:- Estimate the enthalpy of hydrogen iodine formation. Is the reaction endothermic or exothermic? Consider the formation of hydrogen iodide from hydr ogen and iodide: 1 2 𝐻2(𝑔) + 1 2 𝐼2(𝑔) → H𝐼(𝑔) Enthalpy of formation of hydrogen, iodide is the heat change occurring when one atom of hydrogen and one atom of iodine react to form one mole of hydrogen iodide in standard conditions (as gas). To get one atom of hydrogen or iodine, the molecular bond has to be broken. Heat of formation = bond energy of HI - Bond dissociation of 𝐻𝑔 - Bond dissociation energy of 𝐼2 = 299-(218++107) = 299-325 = -26KJ As the heat of formation is negative, the reaction is exothermic.
  • 14. 3. Enthalpy of formation Carbon combines with hydrogen to form many hydrocarbons. Hence, the heat of formation of benzene cannot be determined experimentally but the heat change can be calculated by Hess law 6C + 3𝐻2 → 𝐶 𝐻 , ∆𝐻𝐶6𝐻6 =? Heat of formation of 𝐶𝑂2 and 𝐻2𝑂 are -393.5k and -285KJ respectively. Heat of combustion of benzene is -330KJ such that: C+𝑂2 → C𝑂2 , ∆𝐻1 = -393.5kJ …..(1a) H + 𝑂2 → 𝐻2𝑂 , ∆𝐻2 = -285.8kJ …..(2a) . 𝐶 𝐻 + 9𝑂2 → 6 C𝑂2+3𝐻2𝑂, ∆𝐻3= -3301 kJ (3a) Multiply reaction (i) by 6 6C+6𝑂2 →6C𝑂2, 6∆𝐻1 = -2361kJ (Ib) Multiply reaction (2a) by 3, will give 3𝐻2 + 3𝑂2 → 3𝐻2𝑂, 3∆𝐻2= -857.4kJ (2b)
  • 15. 6C 𝑂2 + 3𝐻2𝑂 → 6C 𝑂2 , 𝐶 𝐻 + 9 𝑂2, −∆𝐻3 = −∆𝐻3 = 3301kJ …(3b) Adding the three reactions give: 6C + 3𝐻2 → 𝐶 𝐻 , ∆𝐻 =82.6kJ The heat of formation of benzene =82.6kJ The heat of formation of benzene = +82.6kJ
  • 16. kb Assignment I The heat of formation of carbon monozide and steam are 111 and 244KJ respectively. Calculate the heat of the reaction: 𝐻2𝑂 + C = CO + 𝐻2 2. Define heat of formation. 3. The reaction between gaseous hydrogen and gaseous Chlorine to form gaseous hydrogen chloride is given as: 𝐻2(g) + Cl(g) 2HCl(g) ΔH = -184.22 kJ . Calculate the heat of formation of Hydrogen Chloride and write the appropriate equation.