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Enthalpy and Hess’s Law

Enthalpy and Hess’s Law

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Enthalpy and Hess’s Law

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  1. Enthalpy and Hess’s Law

  2. From the homework, you may have realized that H can have a negative number. • It relates to the fact that energy as heat has either entered or left the system. • If it was positive, it meant that the heating of the sample required energy. • Energy in means endothermic • So a positiveH means that the process was endothermic. • The opposite is true if H was negative. • A negative H means the process was exothermic (energy left the system)

  3. Thermodynamics • The branch of science that is concerned with the energy changes that accompany chemical and physical changes • To standardize the enthalpies of reactions, scientists have agreed upon a standard thermodynamic temperature of 25.00 C (298.15 K).

  4. Equations and Enthalpy • As a result, we can incorporate more information, like H values and temperatures, into chemical equations. • For example: • Fe (s, 300 K)  Fe (s, 1100K) H = 20.1 kJ/mol • H2 (g, 298 K) + Br2 (l, 298 K)  2 HBr (g, 298K) H = -72.8 kJ/mol

  5. How do they get H values? • To get H values, scientists use calorimetry. • It is defined as the measurement of heat-related constants (specific heat, latent heat) • It is an experimental measurement of an enthalpy change. • A calorimeter is utilized – a device used to measure the heat absorbed or released in a chemical or physical change.

  6. Hess’s Law • It is stated that the amount of heat released or absorbed in a chemical reaction does not depend on the number of steps in the reaction. • Example: Phosphorous pentachloride can be made two ways. • Way #1 – 1 Step • P4 (s) + 10 Cl2 4 PCl5 (g) H = -1596 kJ

  7. Hess’s Law • Way #2 takes two steps • P4 (s) + 6 Cl2 (g)  4 PCl3 (g) H =-1224 kJ • PCl3 (g) + Cl2 (g)  PCl5 (g) H = -93 kJ • Notice that we need to do the second step 4 times to use all the PCl3 from step 1. • So, -1224 +4(-93) = -1596 kJ • Comparing the 1 step way from the last slide to the 2 step way from this slide: • -1596 kJ = -1596 kJ, Hess’s Law works

  8. So, what does Hess’s Law mean? • H reaction = H products - H reactants • So we need to use Standard enthalpy of formation – the enthalpy change in forming one mole of a substance from elements in their standard states. These are listed on Pg 355 and 833. • What these tables mean is when a compound is formed from the elements that make it up, H of formation will be the value listed. • That means that we can find each compound on these lists and use that number in the equation above.

  9. Things to remember • When compounds are multiplied by a constant (coefficient), the enthalpy change must be multiplied by the same constant. So equations will have to be balanced. • H of elements is zero, so those equations may not be in the book. • Remember the diatomics and that S by itself is S8. This may help you find their values.

  10. Now we can look at an example • We want to look at the enthalpy change when carbon and carbon dioxide are reacted together to make carbon monoxide. (C + CO2 2 CO) • So, we need to look at the formation of CO and CO2to complete this problem. • We see that CO, H = -110.5 kJ/mol and CO2, H = -393.5 kJ/mol (C, H = 0) H reaction = H products - H reactants = 2(-110.5)-[(0)+(-393.5)] = 172.5 kJ/mol

  11. Examples • Calculate the enthalpy change for the reaction: 2H2 (g) + 2CO2 (g)  2H2O (g) + 2CO (g) • What enthalpy change accompanies the reaction: 2Al(s) + 3H2O (l)  Al2O3 (s) + 3H2 (g) • Calculate H for the decomposition of calcium carbonate into calcium oxide and carbon dioxide.

  12. Homework • Page 371: 26, 27, 28, 29