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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.

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Presentation on theme: "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."— Presentation transcript:

1 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) + 241.8 KJ Indicates that 1 mol of hydrogen and ½ mol of oxygen produce 1 mol of water The 241.8 KJ represents the enthalpy change per mole of reactant = molar enthalpy

2 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 = - 241.8 KJ/mol to indicate that this is a combustion rxn. See Table 1 on Pg. 306 for a list of molar enthalpies

3 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) + 40.8 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

4 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.

5 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

6 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.

7 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.

8 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


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