1 / 8

Molar Enthalpy

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

brady-rogers
Télécharger la présentation

Molar Enthalpy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Molar Enthalpy • Recall that when we write a thermochemical equation the coefficients represent moles of particles • Therefore, 1 H2(g) + ½ O2(g)  1 H2O(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: ∆Hx • 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, • ∆Hcomb = - 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. H20(l) + 40.8 KJ H2O(g) • Hvap = 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 = nHx • 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 ∆Hsystem = +/- |qsurroundings| 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 = nHx and q = mc∆T • Because of Conservation of Mass, (system) ∆H = q (surroundings = H2O of calorimeter) Therefore, nHx = mc∆T • Note: n and Hx 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 (Tf and Ti) because q has the opposite sign from ∆H

More Related