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Thermochemistry II

Thermochemistry II. Thermochemical equations. Coefficients represent MOLES of reactants and products. Amount of heat energy released or absorbed is proportional to the amounts of reactants and products Allows for the use of fractions as coefficients Phase designations must be included.

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Thermochemistry II

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  1. Thermochemistry II

  2. Thermochemical equations • Coefficients represent MOLES of reactants and products. • Amount of heat energy released or absorbed is proportional to the amounts of reactants and products • Allows for the use of fractions as coefficients • Phase designations must be included

  3. Molar Enthalpy of Formation • ΔH involved in the formation of 1 mole of a compound in its standard state (25°C and 1atm) • ΔH0f indicates standard state enthalpy of formation • ΔH0f values are available in reference materials • Stability of compounds • A negative value indicates product is more stable than reactants.

  4. Molar Enthalpy of Combustion • ΔH involved in the combustion of 1 mole of a reactant in its standard state (25°C and 1atm • This equation is used by every food manufacturer to determine caloric content • A bomb calorimeter measures heat transferred from the combustion reaction to a surrounding water jacket.

  5. Enthalpy Calculations- Hess’s Law • The total ΔH in a reaction is the sum of ΔH for each individual step. • A process can be considered to occur in stages or steps, then the enthalpy change (H) for the overall process can be obtained by summing the enthalpy changes of each individual step • Energy changes are independent of pathway

  6. C(s) + O2(g) (Reactants) DH1=-110.5 kJ CO(g)+1/2 O2(g) DH2=-283 kJ CO2(g) [products] • Overall reaction: C(s) + O2(g) CO2(g) DHtot = DH1 +DH2 = -110.5kJ+(-283kJ) = -393.5 kJ

  7. Determine the Hrx for the following reaction 3C(s) + 4H2(g)  C3H8(g) Hrx =? Given: H (kJ/mol) 1. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) -2045 2. C(s) + O2(g)  CO2(g) -393.5 3. H2(g) + ½ O2(g)  H2O(g) -242

  8. General principles • If a reaction is reversed, then the sign of ΔH must be reversed. • Multiply the coefficients of the known equations. When added we will end up with the desired equation. Multiply ΔH by the same factor.

  9. -1. 3CO2(g) + 4H2O(g) C3H8(g) + 5O2(g) +2045 32. 3C(s) + 3O2(g)  3CO2(g) 3(-393.5) 43. 4H2(g) + 2O2(g)  4H2O(g) 4 (-242) 3C(s) + 4H2(g) C3H8(g) ΔHrx = 2045 kJ+ (-1180.5 kJ) +(-968 kJ) • = -103.5 kJ

  10. Heat of Formation (DHf°) • The standard heat of formation of a compound, DHf°, is the amount of heat absorbed or released when ONE mole of the compound is formed from its elements in their standard state. STANDARD STATE (°) 25°C, 1 atm H2(g) + ½ O2(g) H2O(l)DHf°= -286 kJ/mol H2(g) + ½ O2(g) H2O(g)DHf°= -242 kJ/mol By definition the DHf° of elements in their standard states is equal to ZERO. (O2(g)O2(g)) DHf°(Na(s)) = 0 DHf°(Hg(s))  0, because at 25°C and 1 atm Hg is a liquid DHf°(SO2(g))  0, because SO2 is NOT an element DHf°(I2(g))  0, because at 25°C and 1 atm I2 is a solid

  11. If we know DHf° of the reactants and products in a reaction, we can determine the • DHreaction using the following equation: • DH°reaction=Hf°PRODUCTS - Hf°REACTANTS

  12. Given the following, calculate Hf for the combustion of ethane, C2H6(g) + 7/2 O2(g)  2CO2(g) + 3H2O(l)

  13. DH°reaction= Hf°PRODUCTS - Hf°REACTANTS DH°reaction =sum of[mol of products x Hf°PRODUCTS]- sum of [mol of reactants x Hf°REACTANTS] = [2Hf(CO2(g)) + 3Hf(H2O(l))] – [Hf(C2H6(g)) +7/2 Hf(O2(g))] =[2(-393.5 kJ) +3(-286 kJ)]-[(-84.7 kJ + 7/2(0)] =-1560.3 kJ

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