List of enthalpies for several kinds of reactions. Heats of Reaction List of enthalpies for several kinds of reactions.

How to read the ΔH Table N2 (g) + 3H2 (g)  2NH3 (g) - 91.8 ΔH (kJ) N2 (g) + 3H2 (g)  2NH3 (g) - 91.8 What is the enthalpy for the synthesis of 2 moles of ammonia gas? What is the ∆H for the decomposition of 2 moles of ammonia gas? What is the enthalpy for the decomposition of 1 mole of NH3(g)?

How to read the ΔH Table N2 (g) + 3H2 (g)  2NH3 (g) - 91.8 ΔH (kJ) What is the enthalpy for the synthesis of 2 moles of ammonia gas? -91.8kJ What is the ∆H for the decomposition of 2 moles of ammonia gas? +91.8kJ What is the enthalpy for the decomposition of 1 mole of NH3(g)? + 45.9 kJ

Practice Is the synthesis of aluminum oxide endo or exo? Is the decomposition of nitrogen dioxide endo or exo? How much energy is absorbed if 105.0grams of C2H6 is decomposed?

Practice You MUST use the heats of reaction chart in order to answer these! Is the synthesis of aluminum oxide endo or exo? EXO because -3351kJ Is the decomposition of nitrogen dioxide endo or exo? EXO b/c synthesis is +66.4kJ so the reverse must be the opposite sign. How much energy is absorbed if 105.0grams of C2H6 is decomposed? Convert mass to moles then use the chart. 3.5mol of C2H6 x +84.0kJ = 294kJ

Hess’s Law (of Constant Heat Summation) The energy change of overall reaction (ΔH) is the sum of the individual heats of formation (ΔH0f)

CuO(s) + H2(g) Cu(s) + H2O(g) Practice #1 CuO(s) + H2(g) Cu(s) + H2O(g) Calculate the ΔH given: ΔHof H2O(g) = -242.5 kJ/mol ΔHof CuO(s)= -155.0 kJ/mol Four steps 1. Switch the sign for reactants because they are being broken down NOT formed! 2. Multiply for moles (if necessary)

-87.5 kJ H2O(g) = -242.5 kJ CuO(s)= +155.0 kJ 3. The heat of formation for an element is 0, IGNORE THEM. 4. Add up your numbers H2O(g) = -242.5 kJ CuO(s)= +155.0 kJ -87.5 kJ

Problem #2 CH4 + 3Cl2  CHCl3 + 3HCl The standard heats of formation for CH4, CHCl3 and HCl are -74.8 kJ/mol, -132kJ/mol and -92.0kJ/mol respectively. Calculate the enthalpy for the following: CH4 + 3Cl2  CHCl3 + 3HCl

Problem #2 -333.2kJ CH4 + 3Cl2  CHCl3 + 3HCl The standard heats of formation for CH4, CHCl3 and HCl are -74.8 kJ/mol, -132kJ/mol and -92.0kJ/mol respectively. Calculate the enthalpy for the following: CH4 + 3Cl2  CHCl3 + 3HCl (-74.8) + 3(0) + (-132) + 3(-92.0) = -333.2kJ

Hess’s Law using Elementary steps Calculate ΔH for: C(s) + O2(g) → CO2(g) Given: 2C(s) + O2(g) → 1CO(g)    ΔH= –221.0 kJ CO2(g) → CO(g) + ½O2(g)       ΔH = +283.0 kJ

-393.5 kJ/mol Hess’s Law using Elementary steps Overall: C(s) + O2(g) → CO2(g) ΔH Divide ALL by 2 2C(s) + O2(g)  2CO(g)   –221.0 kJ / 2 C(s) + 0.5 O2(g)   CO(g) 110.5 kJ Flip CO(g) + ½O2(g)  CO2(g)      -283.0 kJ Add together and cancel the same substances on both sides. C(s) + O2(g) + CO(g)  CO(g)  + CO2(g) -393.5 kJ/mol