1 / 18

Entry Task: May 31 st Friday

Entry Task: May 31 st Friday. Turn in Specific Heat Lab. Agenda:. Collect Lab and discuss it Difference between q and ∆H Hess Law notes Hess Law ws. I can…. Use Hess Law of summation of enthalpies ( ∆ H) to calculate the resulting ∆ H of a reaction. Q and ∆H. q=cm ∆T

melita
Download Presentation

Entry Task: May 31 st Friday

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. Entry Task: May 31st Friday Turn in Specific Heat Lab

  2. Agenda: • Collect Lab and discuss it • Difference between q and ∆H • Hess Law notes • Hess Law ws

  3. I can…. • Use Hess Law of summation of enthalpies (∆H) to calculate the resulting ∆H of a reaction

  4. Q and ∆H • q=cm ∆T • Q is the amount heat that a specific amount substance contains- use the specific heat constant to find out the joules of energy. • ∆H is the CHANGE in ENERGY during a reaction or phase change

  5. How come we can’t measure the enthalpy change when Cdiamond (s)  C graphite (s)? • The change takes millions of years • So how can we measure the heat of reaction?

  6. What if I wanted to study the formation of acid rain from the reaction of oxygen (in atmosphere) and gases (sulfur trioxide) from volcanic eruption, how do I do it? • The reaction I’m looking for is: 2S(s) + 3 O2 (g)  2 SO3(g) ΔH = ? • A bit of a problem: When I tried repeating the reaction in the lab, I only produced SO2 • 2S(s) + O2 (g)  SO2(g) ΔH = -297

  7. Hess Law states • That if you can add two or more thermochemical equations to produce a final equation for a reaction. • In other words, take useable pieces from other equations, manipulate them to the equation you want.

  8. Applying Hess law • We want to find out 2S(s) + 3O2(g)  2SO3(g) ΔH = ? We do KNOW and MEASURED: • a. S(s) + O2(g)  SO2 (g) has ΔH of -297 kJ • b. 2SO3(g)  2SO2(g) + O2(g) has ΔH of 198kJ

  9. We want to find out: 2S(s) + 3O2(g)  2SO3(g) ΔH = ? • a. S(s) + O2(g)  SO2 (g) has ΔH of -297 kJ • We can use pieces of each of these equations to create the needed equation from up above. • What pieces of equation a can we use? What do we have to do to equation a to make it useable? • The reactants are great BUT we need 2 times the amount of sulfur. • We have to multiply our ΔH by 2 = -594kJ • 2S(s) + 2O2(g)  2SO2 (g) has ΔH of -594 kJ

  10. We want to find out: 2S(s) + 3O2(g)  2SO3(g) ΔH = ? • b. 2SO3(g)  2SO2(g) + O2(g) has ΔH of 198kJ • We can use pieces of each of these equations to create the needed equation from up above. • What pieces of equation b can we use? What do we have to do to equation b to make it useable? • The reactant needs to be our product • We need to flip equation b so our reactant is now the product –What about the ΔH? • 2SO2(g) + O2(g)  2SO3 (g) ΔH of -198kJ

  11. We want to find out: 2S(s) + 3O2(g)  2SO3(g) ΔH = ? • Now we can add the pieces (Hess Law) of the equations to figure out the ΔH of the desired equation. • Place the NEW a AND b here: Reactant pieces on top • 2S(s) + 2O2(g)  2SO2 (g) ΔH of -594 kJ • 2SO2(g) + O2(g)  2SO3 (g) ΔH of -198kJ • 2S(s) + 3O2(g)  2SO3(g) ΔH = -792 kJ

  12. Use the thermochemical reaction a and b to determine the ΔH for the following decomposition of hydrogen peroxide (H2O2). Desired equation:2H2O2(l)  2H2O(l) + O2 (g) ΔH = ? • a. 2H2(g) + O2 (g)  2H2O(l) ΔH= -527kJ • b. H2(g) + O2(g)  H2O2(l) ΔH= -188kJ

  13. 2H2O2(l)  2H2O(l) + O2 (g) ΔH = ? • a. 2H2(g) + O2 (g)  2H2O(l) ΔH= -527kJ What can we use from equation a? The waters are the products Do we have to modify equation a? Explain. No modification needed because the products are where they need to be

  14. 2H2O2(l)  2H2O(l) + O2 (g) ΔH = ? • b. H2(g) + O2(g)  H2O2(l) ΔH= -188kJ What can we use from equation b? The H2O2 Do we have to modify equation b? Explain. Yes, we need to flip the equation so H2O2 is on reactant side, change the ΔH sign AND multiply it by 2 2H2O2(l) + 2H2(g)  2O2(g) ∆H=376 kJ

  15. 2H2O2(l)  2H2O(l) + O2 (g) ΔH = ? Place the NEW a AND b here: Reactant pieces on top ________  ________ + ________ ΔH= _________ ________ + ________  ________ ΔH= _________ 2H2O2(l) 2H2(g) 2O2(g) 376 kJ 2H2(g) O2 (g) 2H2O(l) -527 kJ COMBINE ALL reactants 2H2O2(l)  2H2O(l) + O2 (g) ΔH = -151 kJ

  16. Desired: 2CO(g) + 2NO (g)  2CO2(g) + N2(g) ΔH= ? a. 2CO(g) + O2 (g)  2CO2(g) ΔH = -566.0 kJ b. N2(g) + O2(g)  2NO(g) ΔH= 180.6kJ 2CO(g) + O2 (g)  2CO2(g) ΔH = -566.0 kJ b. 2NO(g)  O2(g) + N2(g) ΔH= -180.6kJ 2CO(g) + 2NO (g)  2CO2(g) + N2(g) ΔH= -746.6kJ

  17. Desired: 4Al(s) + 3MnO2(s) 2Al2O3(s) + 3Mn(s) ΔH= ? a. 4Al(s) + 3O2 (g)  2Al2O3 (s) ΔH = -3352 kJ b. Mn(s) + O2(g)  MnO2(s) ΔH= -521 kJ • 4Al(s) + 3O2 (g)  2Al2O3 (s) ΔH = -3352 kJ b. 3MnO2(s)  3Mn(s) + 3O2(g) ΔH= 1563 kJ 4Al(s) + 3MnO2(s)  2Al2O3(s) + 3Mn(s) ΔH= -1789kJ

  18. Next Time:Hess’s law worksheet

More Related