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For exothermic rxns, the heat content of the reactants is larger than that of the products. enthalpy of reaction:  H rxn = H products – H reactants (also.

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Presentation on theme: "For exothermic rxns, the heat content of the reactants is larger than that of the products. enthalpy of reaction:  H rxn = H products – H reactants (also."— Presentation transcript:

1 For exothermic rxns, the heat content of the reactants is larger than that of the products. enthalpy of reaction:  H rxn = H products – H reactants (also called “heat of reaction”)

2  H = –2390 kJ 2 H 2 (g) + O 2 (g)  2 H 2 O(g)  H = –483.6 kJ What is the enthalpy change when 178 g of H 2 O are produced? 178 g H 2 O The space shuttle was powered by the reaction above.

3  H for a reaction and its reverse are the opposites of each other. Enthalpy change depends on the states of reactants and products. 2 H 2 (g) + O 2 (g)2 H 2 O(g)(  H = –483.6 kJ) 2 H 2 O(g) 2 H 2 (g) + O 2 (g)(  H = +483.6 kJ) Enthalpy/energy is a reactant. 2 H 2 (g) + O 2 (g)2 H 2 O(g)(  H = –483.6 kJ) 2 H 2 (g) + O 2 (g)2 H 2 O(l)(  H = –571.6 kJ)

4 Calorimetry: the measurement of heat flow -- device used is called a...calorimeter heat capacity of an object: amount of heat needed to raise object’s temp. 1 K = 1 o C molar heat capacity: amt. of heat needed to raise temp. of 1 mol of a substance 1 K specific heat (capacity): amt. of heat needed to raise temp. of 1 g of a substance 1 K i.e., molar heat capacity = molar mass X specific heat

5 c X = heat of fusion (s/l) or heat of vaporization (l/g) We calculate the heat a substance loses or gains using: where q = heat m = amount of substance c P = substance’s heat capacity  T = temperature change q = m c P  T (for within a given state of matter) AND q = + / – m c X (for between two states of matter)

6 HEAT Temp. s s/l l l/g g heat added (+q)   heat removed (–q) Typical Heating Curve

7 What is the enthalpy change when 679 g of water at 27.4 o C are converted into water vapor at 121.2 o C? HEAT Temp. s s/l l l/g g q = m c P  T c f = 333 J/g c v = 40.61 kJ/mol c P,l = 4.18 J/g-K c P,s = 2.077 J/g-K c P,g = 36.76 J/mol-K Heat liquid… = 679 g (4.18 J / g-K ) (100 – 27.4)= 206 kJ q = +m c X Boil liquid… = +37.72 mol (40.61 kJ / mol ) = 1532 kJ q = m c P  T Heat gas… = 37.72 mol (36.76 J / mol-K ) (121.2–100)= 29.4 kJ  H = 1767 kJ +

8 With a coffee-cup calorimeter, a reaction is carried out under constant pressure conditions. -- Why is the pressure constant? calorimeter isn’t sealed, atmospheric pressure is constant -- If we assume that no heat is exchanged between the system and the surroundings, then the solution must absorb any heat given off by the reaction. -- For dilute aqueous solutions, it is a safe assumption that c P = 4.18 J/g-K the specific heat of water i.e., q absorbed = –q released


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