1 / 7

STANDARD HEAT OF FORMATION ΔH 0 f or ΔH θ f

STANDARD HEAT OF FORMATION ΔH 0 f or ΔH θ f. It is define as the change in enthalpy that accompanies the formation of one mole of a compound from its elements at standard states of 25 0 C and 101.3kPa. What are the standard conditions?. ΔH 0 f or ΔH θ f.

shiloh
Download Presentation

STANDARD HEAT OF FORMATION ΔH 0 f or ΔH θ f

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. STANDARD HEAT OF FORMATIONΔH0f or ΔHθf It is define as the change in enthalpy that accompanies the formation of one mole of a compound from its elements at standard states of 250C and 101.3kPa. What are the standard conditions?

  2. ΔH0f or ΔHθf ΔH means enthalpy changes or heat of content 0 or θ means standard conditions: 250C and 101.3kPa f means formation Please open to page 530, Table 17.4

  3. Enthalpy changes of formation are useful for working out enthalpy changes you can’t find directly. You need to know the ΔH0f for all the reactants and products that are compounds. The ΔH0f for elements or diatomic molecules is zero – the element is being formed from the element so there’s no change.

  4. REACTANTS PRODUCTS ΔH0r SO2(g) + 2H2S(g) 3S(s) + 2H2O(l) Sum all ΔH0f(products) Sum all ΔH0f(reactants) ΔH0r = - ΔH0f(reactants) ΔH0f(products)

  5. EXAMPLE Find the ΔH0r SO2(g) + 2H2S(g) 3S(s) + 2H2O(l) ΔH0f[SO2(g)] = -297 kJmol-1 ΔH0f[H2S(g)] = -20.2 kJmol-1 ΔH0f[H2O(l)] = -286kJmol-1

  6. ΔH0r = - ΔH0f(reactants) ΔH0f(products) To find ΔH0r of this reaction; SO2(g) + 2H2S(g)3S(s) + 2H2O(l) REACTANTS PRODUCTS ΔH0r = [0 + (-286 x 2)] – [-297 + (-20.2 x 2)] ΔH0f of sulphur is 0 because it’s an element There’s 2 moles of H2O and 2 moles of H2S

  7. ΔH0r = [0 + (-286 x 2)] – [-297 + (-20.2 x 2)] = -234.6 kJmol-1

More Related