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Enthalpy and Hess’s Law
Chemistry I
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Chemical Reactions at Constant Volume and Pressure
We know DE = q + w DE = q – PDV At constant volume, PDV = 0 So at constant volume, DE = qv Subscript “v” reminds us that is at constant volume
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Chemical Reactions at Constant Pressure
This equation already tells us at constant pressure, DE = qp – PDV We can rearrange the equation as: qp = DE – PDV We can redefine “q” as H, or enthalpy H = E + PV If pressure is held constant, DH = DE + PDV H is also a state function Hint: most uppercase symbols are state functions
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Enthalpy Enthalpy – heat absorbed or released in a reaction
Depends on the difference in quantity of heat Represented by a H When the pressure of a reaction remains constant, the heat absorbed or released during a chemical reaction is equal to the change in enthalpy for the reaction
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Enthalpy Change We can determine how energy was changed throughout a reaction DH = Hproducts – Hreactants Endothermic – Heat absorbed ends up as “extra” enthalpy for the products compared to the reactants DH = + (positive) Exothermic – Heat absorbed ends up as “less than” the enthalpy for the products compared to the reactants DH = - (negative)
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Enthalpy Diagram Exothermic reaction
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Enthalpy Sign of DH Process Heat Positive Endothermic Absorbed
Negative Exothermic released
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Standard Enthalpy Enthalpy measured when reactants are in their standard state (how they exist at room temperature) they are referred to as standard enthalpy DH⁰
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Using Enthalpy Changes
How much heat is released if 1.0 g of H2O2 decomposes in a beetle to produce a steam spray? 2 H2O2(l) 2 H2O (l) + O2(g) DH⁰ = -190 kJ 1.0 g H2O2 x (1 mol H2O2 / 34.0 g H2O2) = mol H2O2 Heat transferred = mol H2O2 x (-190 kJ/2 mol H2O2) = -2.8 kJ
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Using Enthalpy Changes
How much heat is released if 6.44 g of S reacts with O2 according to the following equation? 2S +3 O2 2 SO3 DH⁰ = kJ 6.44 g S x (1 mol S / 32.1 g S) = mol S Heat transferred = mol S x ( kJ/2 mol 3 O2) = 79.5 kJ
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Relationship between DH and DE
Under constant volume, DE =qv Under constant pressure, DH = qp For most reactions not involving gases, DE is almost the same as DH since there (since there is no change in DV) For gases, some of the energy (DE) is used to expand (or contract) the gas, therefore DH will be slightly smaller (DV is changing)
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Relationship Cont. For gases, therefore, DE = DH – D(PV)
From ideal gas: PV = nRT So DE = DH – RTDn Dn = number of moles of product GAS – number of moles of reactant GAS
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Example 6.4 Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25C. 2 CO(g) + O2(g) 2 CO2(g) DH = kJ/mol DE = DH – RTDn Dn = 2 -3 = -1 DE = kJ/mol – (8.314 J/K*mol)(1 kJ/1000 J)(298 K)(-1) kJ/mol
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