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Tuesday 7 February 2012

Energy in the Equation & Energy Calculations

Energy in the Equation:

  • Energy absorption or release can be placed directly in equation
    • For example (the formation of water): 2H2 (g) + O2 (g) --> 2H2O (l) + 572 kJ
  • Exothermic reactions have the energy term on the right hand side and a negative change in enthalpy (H).  Therefore the formation of water is an exothermic reaction
  • Endothermic reactions have the energy term on the left hand side and a positive change in enthalpy (H).  
Energy Calculations:
  • Change in enthalpy is the energy change of a reaction and is expressed in kJ/mole of 1 of the chemicals
  • We take the water formation sample again: 2H2 (g) + O2 (g) --> 2H2O (l) + 572 kJ
    • Change in enthalpy for the above reaction is expressed using coefficients of the balanced equation:
      • -572 kJ / 1 mol O2 or
      • -572 kJ / 2 mol H2 = -286 kJ / 1 mol H2 or
      • -572 kJ / 2 mol H2O = -286 kJ / 1 mol H2O
    • Therefore, the value of change in enthalpy depends on which chemical in the chemical equation is referred to

Again, an example for the exothermic reaction:


Note:
  1. The value of change in enthalpy with different reactions
  2. change in enthalpy depends on the chemical reaction
  3. change in enthalpy is NOT a constant like Avogadro's number

Now, we officially welcome the return of moles and significant figures!!!

Sample question you can be asked:

e.g. For the reaction 2H2 (g) + O2 (g) --> 2H2O (l) + 572 kJ, if 2.3 moles of oxygen gas is burned in the presence of hydrogen gas, how much heat energy would be released?

2.3 mol O2 * (-572 kJ/1 mol O2) = 1315.6 kJ

Don't forget sig. figs: = 1300 kJ

Relationship between Moles and energy:

Moles * (kJ/1mol) = Energy
Energy * (1 mol/kJ) = Moles

Links:
(Enthalpies of reactions)
(Enthalpy changes)
(Practice problems with answers)


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