September 01, 2015 By the end of today's lesson you should · know what enthalpy is · · · understand the difference between a molar enthalpy value and an enthalpy change · · · · be able to use a formula to calculate an enthalpy change September 01, 2015 Thermal energy is released or absorbed by chemical reactions Endothermic Reactions · more energy is required to break bonds than is released by bond formation Exothermic Reactions · more energy is released in bond formation that is required to break bonds September 01, 2015 September 01, 2015 Each chemical reaction has its own enthalpy change ∆H depends on: -the reaction (products and reactants) -the coefficients of the balanced equation Example N2(g) + 3 H2(g) --> 2 NH3(g) ∆H = -90 kJ 2 N2(g) + 6 H2(g) --> 4 NH3(g) ∆H = September 01, 2015 September 01, 2015 Example: The combustion of ethane (C2H6(g)) 2 C2H6(g) + 5 O2(g) ---> 2 CO2(g) + 6 H2O(g) ∆H =-3122 kJ What is the molar enthalpy of reaction for ethane? What is the molar enthalpy of reaction for water? An Analogy.... A family of four wants to take a plane ride from Edmonton to Vancouver. Each ticket costs $200. The total cost of the trip for the family will be $800. In this analogy what would be the chemical reaction molar enthalpy of reaction enthalpy change September 01, 2015 Converting from an enthalpy change to a molar enthalpy value Example 2 Example 1 2 NaOH(s) + 2 Al(s)+ 2 H2O(l) --> 2NaAlO2(aq) + 3 H2(g) 2 KCl(s) + 2 H2SO4(l) ----> 2 HCl(g) + K2SO4(s) ∆H = -850 kJ ∆H = + 41.0 kJ What is the molar enthalpy of reaction for hydrogen? What is the molar enthalpy change for HCl(g)? Converting from a molar enthalpy to an enthalpy change September 01, 2015 Converting from a molar enthalpy to an enthalpy change ∆H = n ∆rH Example1 Example 2 SO + 2 H S --> 3 S + 2 H O 2 ZnS(s) + 3 O2(g) ----> 2 ZnO(s)+ 2 SO2(g) 2(g) 2 (s) (l) 2 (g) ∆ H (S) = -31.0 kJ/mol ∆rH (ZnS) = - 439.1 kJ/mol r What is the enthalpy change for the reaction? What is the enthalpy change for this reaction? ∆H = n∆rH ∆H = 3 mol S x -31.0 kJ/mol ∆H = -93.0 kJ Converting from one molar enthalpy value to another given a balanced equation September 01, 2015 Converting from one molar enthalpy value to another given a balanced equation step 1: Find the enthalpy change for the reaction step 2: Find the new molar enthalpy value Example 1 Example 2 SO + 2 H S --> 3 S + 2 H O 2(g) 2 (s) (l) 2 (g) 2 ZnS(s) + 3 O2(g) ----> 2 ZnO(s)+ 2 SO2(g) ∆ H (S) = -31.0 kJ/mol ∆ H (ZnS) = - 439.1 kJ/mol r r What is the molar enthalpy of reaction for H S ? What is the molar enthalpy of reaction for O ? 2 (g) 2(g) Step 1: ∆H = n∆rH ∆H = 2 mol ZnS x -439.1 kJ/mol ∆H = -878.2 kJ Step 2: ∆rH = ∆H /n ∆rH (O2) = -878.2 kJ / 3 mol O2 ∆rH (o2) = -292.7 kJ Stoichiometry and Enthalpy Changes The amount of energy released from a chemical reaction is affected to the number of moles of reactant or product. If given the mass of reactant or product, the enthalpy change (∆H in kJ) can be determined from the molar enthalpy value for the reactant or product. September 01, 2015 Example 1 Example 2 C H O + 12 O ---> 12 CO + 11 H O 12 22 11(s) 2(g) 2(g) 2 (l) SO2(g) + 2 H2S(s) --> 3 S(l) + 2 H2O(g) ∆rH(S) = - 31.0 kJ/mol What is the enthalpy change for the reaction of 100 g of S(s) ? ∆rH (C12H22O11) = -5640.3 kJ/mol What is the enthalpy change for the reaction of 20.0 g of sucrose (C12H22O11)? Calculating an enthalpy change for a given mass of reactant or product given a balanced equation with an enthalpy change step 1: Find the molar enthalpy for the substance you are given the mass of step 2: Find the ∆H = n ∆rH, where n=m/M Example 1 Example 2 2 NaOH + 2 Al + 2 H O --> 2NaAlO + 3 H (s) (s) 2 (l) 2(aq) 2(g) 2 KCl(s) + 2 H2SO4(l) ----> 2 HCl(g) + K2SO4(s) ∆H = -850 kJ ∆H = + 41.0 kJ How much energy is releaseD when 250 grams of How much energy is requireD to proDuce 40.0 grams of aluminum reacts? K2SO4? Step 1: ∆rH = ∆H/n ∆H = +41.0 kJ/ 1 mol K2SO4 ∆H = +41.0 kJ/mol Step 2: ∆H = n ∆r H = 40.0 g/174.27 g/mol x +41.0 kJ/mol ∆H = +9.41 kJ -3.94 x 103 kJ September 01, 2015 Determining moles of reactant or product given an enthalpy change Example 1 Example 2 C H O + 12 O ---> 12 CO + 11 H O 12 22 11(s) 2(g) 2(g) 2 (l) 2 KCl(s) + 2 H2SO4(l) ----> 2 HCl(g) + K2SO4(s) ∆rH (C12H22O11) = -5640.3 kJ/mol ∆H = + 41.0 kJ If 10 000 kJ of energy is releaseD, what mass of sucrose reacteD? If 300 kJ of energy is absorbed, how much KCl(s) reacted? Step 1: Find ∆rH for KCl ∆rH = ∆H/n = +41.0 kJ/ 2 mol KCl ∆rH = +20.5 kJ/mol n = ∆H / ∆rH n = + 300 kJ / + 20.5 kJ/mol n= 14.6 mol m = 14.6 mol x 74.55 g/mol = 1076 g m = 1.09 kg.