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Entropy & Chemical Reactions

Entropy & Chemical Reactions. 2 nd Law of Thermodynamics. A process will be spontaneous is the entropy of the universe increases Now we will look at entropy regarding to chemical reactions. What is the sign for S?. N 2 (g) + 3H 2 (g)  2NH 3 (g)

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Entropy & Chemical Reactions

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  1. Entropy & Chemical Reactions

  2. 2nd Law of Thermodynamics • A process will be spontaneous is the entropy of the universe increases • Now we will look at entropy regarding to chemical reactions

  3. What is the sign for S? • N2(g) + 3H2(g)  2NH3(g) • First look at the states…if you go from a solid or a liquid to a gas, you will have a + entropy • In this case, all of the states are the same, so we look at the number of moles • N2(g) + 3H2(g)  2NH3(g) • 1 + 3 vs. 2 • The entropy decreases because you go from 4 moles to 2 • S is negative

  4. What is the sign for S? • 4NH3 + 5O2 4NO + 6H2O • 9 moles vs. 10 moles • S increases (+)

  5. Calculating S • Calculating S is just like calculating H • Simply use the Appendix…just look at the column for S instead of H • S° of any element or diatomic molecule is NOT zero. • You must look these up!

  6. Example • Calculate S for the following reaction: • 2NiS(s) + 3O2(g)  2SO2(g) + 2NiO (s) • 2(-53) + 3(-205) + 2(248) + 2(38) • -149 J/K

  7. Example • Calculate S for the following reaction: • Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g) • (-51) + 3(-131) + 2(28) + 3(189) • 179 J/K

  8. Gibbs Free Energy & Chemical Reactions • You can calculate G in 3 ways… • Like Hess’s Law • Like H° • With the equation G = H - T S

  9. Example • Calculate H, S, & G at 25°C using the following data… • 2SO2 + O2 2SO3

  10. Example 2SO2 + O2 2SO3 H = 2(+297)+(0) + 2(-396) = -198 KJ S = 2(-248)+ (-205) + 2(257) = -187J/K G = H – T S =(-198) – (298 x -0.187) =-142 KJ

  11. Calculate G • Using the following data at 25°C • Cdiamond + O2(g)  CO2 (g) G = -397KJ • Cgraphite + O2(g)  CO2 (g) G = -394KJ • Calculate G for the reaction: • C diamond C graphite • G = -3KJ

  12. Calculating G • Methanol is a high octane fuel used in high performance racing engines. Calculate G for the following reaction • 2 CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g) • Given the following free energies of formation:

  13. Calculating G • 2 CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g) • 2(163) + 3(0) + 2(-394) + 4(-299) • -1378 KJ

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