Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chemistry 100 Enthalpy and Hess’s Law. Energy Changes in Chemical Reactions Let’s take a typical reaction CH 4 (g) + O 2 (g)  CO 2 (g) + 2 H 2 O (l)

Similar presentations


Presentation on theme: "Chemistry 100 Enthalpy and Hess’s Law. Energy Changes in Chemical Reactions Let’s take a typical reaction CH 4 (g) + O 2 (g)  CO 2 (g) + 2 H 2 O (l)"— Presentation transcript:

1 Chemistry 100 Enthalpy and Hess’s Law

2 Energy Changes in Chemical Reactions Let’s take a typical reaction CH 4 (g) + O 2 (g)  CO 2 (g) + 2 H 2 O (l) This is an example of ‘burning a fossil fuel.’ This reaction releases energy! What about this reaction? KCl (s)  KCl (aq) This reaction requires energy!

3 Energy Changes Almost all chemical and physical processes are accompanied by the loss or gain of energy from the system. System energy surroundings

4 Some Definitions Exothermic reaction  heat is released from the system to the surroundings  e.g., the combustion of methane CH 4 (g) + O 2 (g)  CO 2 (g) + 2 H 2 O (l) Endothermic reaction  heat is supplied to the system by the surroundings.  e.g., dissolving potassium chloride in water. KCl (s)  KCl (aq)

5 Enthalpy Enthalpy H  the heat content of a substance. Enthalpy change -  H thermal energy (heat) evolved or absorbed under constant pressure conditions).

6 Exothermic vs. Endothermic Processes  Exothermic process  heat content of system decreases, the system enthalpy decreases (i.e.,  H < 0)  Endothermic process  energy of the system is increased (i.e.  H > 0).

7 The Enthalpy Change in a Chemical Reaction We want to know how much energy is released when we burn ethane! C 2 H 6 (g) + 7/2 O 2 (g)  2 CO 2 (g) + 3 H 2 O (l) We wish to know the difference in the heat content of the reactants vs. the products.   H   n p H (products) -  n r H (reactants)  n p and n r represent the number of moles of products and reactants, respectively.

8 The Enthalpy Change (cont’d) For the ethane combustion reaction 1 C 2 H 6 (g) + 7/2 O 2 (g)  2 CO 2 (g) + 3 H 2 O (l)  H   n p H (products) -  n r H (reactants) = 3 H[H 2 O (l)] + 2 H[CO 2 (g)] - (7/2 H[O 2 (g)] + 1 H[C 2 H 6 (g)] )

9 Conventions for Writing Thermochemical Equations For exothermic reactions,  H 0.  H values are given for processes occurring at 25  C and 1 atm pressure  STTP - standard thermodynamic temperature and pressure.

10 Conventions (cont’d) The physical state of each substance participating in the process must be stated  s  solid  l  liquid  g  gas  aq  aqueous solutions

11 Conventions (cont’d) The stoichiometric coefficients = the number of moles of each substance involved in the transformation Fractional stoichiometric coefficients are permitted in thermochemical equations.

12 Conventions (cont’d) Multiply or divide a chemical equation by a factor, the  H value must also be multiplied or divided by that factor  Enthalpy is an extensive property. When an equation is reversed, the sign of the  H value is changed but its magnitude stays the same.

13  H Values and Hess’s Law Calculating enthalpy changes for physical and chemical transformations.  tabulated values of reaction enthalpies (largely measured with a calorimeter).  indirect Method – Hess’s Law.

14 Hess’s Law  Hess’s Law  enthalpy changes for sequences of reactions.  Enthalpy change accompanying a reaction is the same whether the reaction occurs in a single step or in many steps.


Download ppt "Chemistry 100 Enthalpy and Hess’s Law. Energy Changes in Chemical Reactions Let’s take a typical reaction CH 4 (g) + O 2 (g)  CO 2 (g) + 2 H 2 O (l)"

Similar presentations


Ads by Google