Download presentation
Presentation is loading. Please wait.
Published byEsther Reed Modified over 9 years ago
1
A Comparison of H and E 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2 (g) H = -367.5 kJ/mol E = H - P V At 25 0 C, 1 mole H 2 = 24.5 L at 1 atm P V = 1 atm x 24.5 L = 2.5 kJ E = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol 6.4
2
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = ms t q = C t t = t final - t initial 6.5
3
How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = 0.444 J/g 0 C t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = ms t = 869 g x 0.444 J/g 0 C x –89 0 C= -34,000 J 6.5
4
Constant-Volume Calorimetry No heat enters or leaves! q sys = q water + q bomb + q rxn q sys = 0 q rxn = - (q water + q bomb ) q water = ms t q bomb = C bomb t 6.5 Reaction at Constant V H ~ q rxn H = q rxn
5
Constant-Pressure Calorimetry No heat enters or leaves! q sys = q water + q cal + q rxn q sys = 0 q rxn = - (q water + q cal ) q water = ms t q cal = C cal t 6.5 Reaction at Constant P H = q rxn
6
6.5
7
Chemistry in Action: Fuel Values of Foods and Other Substances C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) H = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J
8
Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation ( H 0 ) as a reference point for all enthalpy expressions. f Standard enthalpy of formation ( H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero. H 0 (O 2 ) = 0 f H 0 (O 3 ) = 142 kJ/mol f H 0 (C, graphite) = 0 f H 0 (C, diamond) = 1.90 kJ/mol f 6.6
10
The standard enthalpy of reaction ( H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d H 0 (D) f c H 0 (C) f = [+] - b H 0 (B) f a H 0 (A) f [+] H0H0 rxn n H 0 (products) f = m H 0 (reactants) f - 6.6 Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
11
Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g) H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g) H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) H 0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g) H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g) H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l) H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 6.6
12
Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n H 0 (products) f = m H 0 (reactants) f - H0H0 rxn 6 H 0 (H 2 O) f 12 H 0 (CO 2 ) f = [+] - 2 H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6535 kJ -6535 kJ 2 mol = - 3267 kJ/mol C 6 H 6 6.6
13
The enthalpy of solution ( H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. H soln = H soln - H components 6.7 Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack?
14
The Solution Process for NaCl H soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol 6.7
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.