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CHEMICAL REACTIONS AND ENERGY - 2

CHEMICAL REACTIONS AND ENERGY - 2. ENTHALPY (H). Accounts for heat flow in chemical reactions happening at constant pressure when work is performed due to P-V change . The enthalpy of a substance is its internal energy plus a term that takes into account the pressure & volume of the substance.

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CHEMICAL REACTIONS AND ENERGY - 2

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  1. CHEMICAL REACTIONS AND ENERGY - 2

  2. ENTHALPY (H) • Accounts for heat flow in chemical reactions happening at constant pressure when work is performed due to P-V change. • The enthalpy of a substance is its internal energy plus a term that takes into account the pressure & volume of the substance. H = U + PV

  3. We cannot calculate the actual value of H. Instead, we can calculate the change in enthalpy. ΔH = ΔU + PΔV (since W = - PΔV) • (Equation 1) ΔH = ΔU – W In a process carried out at constant pressure: ΔU = qp + W or (Equation 2) (Equation 1) = (Equation 2) ΔH = qp (at constant pressure) qp= ΔU - W

  4. ONLY at constant Pressure: ΔH = Qp For this reason, the term “ heat of reaction” and “change in enthalpy” are used interchangeably for rxns studied at constant pressure.

  5. In summary At constant volume At constant pressure • ΔU measures heat lost/gained. • ΔH measures heat lost/gained. The difference between ΔU & ΔH is very small. ΔH is generally satisfactory to use.

  6. Exercise • Under standard conditions, 1 mol CO is burnt in a sealed flask w/ constant volume and 281.75 kJ energy is released. The same amount of CO is burnt under the same conditions in a flask w/ frictionless movable piston, 283kJ of energy is released. What are the values of ΔH, ΔU, and w for both conditions?

  7. Solution: • 1st condition--constant volume, w=0 ΔU=qv = -281.75 kJ ΔH=ΔU-w (w=0)- ΔH= ΔU=-281.75 kJ, w=0 2nd condition--constant pressure, ΔH= qp= -283kJ ΔU=-281.75 kJ, ΔH=ΔU-w, w=ΔU-ΔH, w= - 281.75 + 283kJ= + 1.25 kJ

  8. For a chemical rxn: ΔH = Hproducts–Hreactants

  9. WhenHproducts> Hreactants • ΔH is (+). • Heat will be absorbed by the system. • Reactants are more stable than the products. • reactants have stronger bonds than the products.

  10. WhenHproducts< Hreactants • ΔH is (-). • Heat will be released by the system. • Reactants are less stable than the products. • Products have stronger bonds than the reactants.

  11. ENTHALPIES OF RXNS • Enthalpy is an extensive property. ΔH is directly proportional to the amounts of reactants consumed in chemical rxns. CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH = -890 kJ 2CH4(g) + 4O2(g) --->2 CO2(g) + 4H2O(l) ΔH = -1780 kJ (890x2)

  12. Exercise • When 1 mole of methane is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which a 5.8 g sample of methane is burned at constant pressure.

  13. Solution: • CH4: 16 g/mol • When 16g of CH4 burns 890 kJ energy is released 5.8 g of CH4 ? ________________________________________ ? = - 320 kJ of energy

  14. 2) ΔH for a rxn is equal in magnitude but opposite in sign, to ΔH for the reverse rxn. CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH = -890 kJ CO2(g) + 2H2O(l) ---> CH4(g) + 2O2(g) ΔH = + 890 kJ

  15. 3) ΔH depends on the state (gas, liquid, solid, crystalline structure)of the reactants and products. • Therefore, we should write the states of thesubstances in the rxn equation. • I. CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH1 = -890 kJ • II. CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(g) ΔH2 = ? kJ • Which one is greater in amount?

  16. ΔH1 ( -890 kJ)> ΔH2 (- 802 kJ): H2O(l)  H2O(g)ΔH= + 88 kJ

  17. 4) ΔH depends on temperature & pressure of the rxn medium. - We will generally assume that the reactants & products are both at the same temperature, 25 ° C, unless otherwise stated.

  18. 5) ΔH is a state function. Therefore, it doesn’t depend on the rxn pathway.

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