Molar Enthalpy Recall that when we write a thermochemical equation the coefficients represent moles of particles Therefore, 1 H 2(g) + ½ O 2(g)  1 H 2.

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Molar Enthalpy Recall that when we write a thermochemical equation the coefficients represent moles of particles Therefore, 1 H 2(g) + ½ O 2(g)  1 H 2 O (g) KJ Indicates that 1 mol of hydrogen and ½ mol of oxygen produce 1 mol of water The KJ represents the enthalpy change per mole of reactant = molar enthalpy

Representing Molar Enthalpies We use the following symbol to show molar enthalpy: ∆H x X is a letter or series of letters that indicate the type of change that is occurring For the reaction on the previous slide we use, ∆H comb = KJ/mol to indicate that this is a combustion rxn. See Table 1 on Pg. 306 for a list of molar enthalpies

Why use Molar Enthalpy? Allows us to describe the energy change in a given chemical reaction/mol of substance. Also used for physical changes such as changes of state. H 2 0 (l) KJ  H 2 O (g) H vap = 40.8 KJ/mol – this represents the change in potential energy in the system See Table 2 – Pg. 307

Calculations Involving Molar Enthalpies To calculate the amount of energy involved in a particular change (∆H) we need to multiply the molar enthalpy by the number of moles ∆H = nH x Where, n = number of moles Look at the sample problem on Pg. 307 Complete Pg. 308 # 1-3.

Calorimetry Based on the Law of Conservation of Energy ∆H system = +/- |q surroundings | Therefore, we can measure the total energy change of a system by measuring the total energy change of the surroundings Calorimetry involves measuring energy changes in a closed container called a calorimeter

Assumptions of Calorimetry No heat is transferred b/n the calorimeter and the environment Any heat absorbed or released by the calorimeter itself is negligible (does not affect the result) A dilute aqueous solution is assumed to have a density and specific heat capacity of water See diagram on Pg. 309.

Calculations with Calorimetry – Finding Molar Enthalpies Recall: ∆H = nH x and q = mc∆T Because of Conservation of Mass, (system)∆H = q (surroundings = H 2 O of calorimeter) Therefore, nH x = mc∆T Note: n and H x refer to the solute m, c and ΔT refer to the solvent and are assumed to have the same density and specific heat capacity as water.

Practicing Calorimetry Calculations See sample problem on Pg. 309 Complete Practice Problems on Pg. 310 # 4 and 5. Note: The temperature terms are reversed here (T f and T i ) because q has the opposite sign from ∆H