Kj mol 1 2 hesss law 3 2 h 2 g o2 g

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Unformatted text preview: O(g) ∆H = −483.9kJ / mol (2) 2 H 2O ( g ) → 2 H 2O (l ) ∆H = −88kJ / mol (3) 2 H 2 ( g ) + O2 ( g ) → 2 H 2O(l ) ∆H = ? kJ / mol (1) +(2) Hess’s Law (3) 2 H 2 ( g ) + O2 ( g ) = 2H 2 O(g) 2 H 2O( g ) = 2 H 2O (l ) 2 H 2 ( g ) + O2 ( g ) = 2 H 2O (l ) ∆H 3 = ∆H1 + ∆H 2 = −438.9kJ − 88kJ = −572kJ ∆H of overall reaction is the sum of the ∆H for each sum ∆H = ∆H1 + ∆H 2 + ∆H 3 individual step Exercise: Given the changes of enthalpy of reaction (1) & (2). Calculate how much heat is involved in the 3rd reaction? N2(g) + 2O2(g) → 2NO2(g) 2 NO (g) + O2(g) → 2NO2(g) N2(g) + O2(g) → 2NO(g) N 2 (g) + 2O 2 (g) → 2NO 2 (g) + 2NO 2 (g) → 2 NO (g) + O 2 (g) N 2 (g) + O 2 (g) → 2NO(g) ∆H1 = 66.4 kJ ∆H2 = - 114.2 kJ ∆H3 = ? ∆H1 = 66.4 kJ ∆H 2 = +114.2 kJ ∆H 3 =(66.4+114.2)kJ Hess ' s Law : ∆H = ∆H1 + ∆H 2 + ∆H 3 Exercise: Given the changes of enthalpy of reaction (1), (2) & (3) Calculate how much heat is involved in the 4th reaction? NO(g) NO(g) + O3(g) → NO2(g) + O2(g) ∆H1 = -198.9 kJ 2 O3(g) → 3 O2(g) ∆H2 = -284.6 kJ O2(g) → 2O(g) ∆H3 = 495.0 kJ NO(g) + O(g) → NO2(g) ∆H4= ? NO(g) + O3 (g) → NO 2 (g) + O 2 (g) ∆H1 = -198.9 kJ 3 O (g) → O (g) 3 22 O(g) → 1 O 2 (g) 2 ∆H '2 = - (-284.6) 2 kJ ∆H '3 = - (495.0) 2 kJ NO(g) + O(g) → NO 2 (g) ∆H 4 = -198.9 kJ- (-284.6) 2 kJ- (495.0) 2 kJ = -304.1kJ Enthalpy Enthalpy and the First Law of Thermodynamics The change of enthalpy ( H) equals the heat (qp) gained or lost by the system when the process occur at the constant pressure ∆E = q + w q p = ∆H At constant Presure: w = - P∆V ∆E = ∆H − P ∆ V ∆H = ∆E + P∆V = q P ∆H = ∆E + P∆V = q P Measurement ---Calorimetry Measurement ---Calorimetry • Calorimetry : A careful measurement of heat flow • Calorimeter : apparatus that measures heat flow apparatus ConstantConstant-Pressure Calorimeter Calorimeter A well insulated compartment to capture capture heat and house the reactants reactants ConstantConstant-Volume Calorimeter Calorimeter Calorimetry Sealed, air tight when volume is constant, constant, V=0 w = - P∆V = 0 ∆E v = qv + w = qv Measure qv (or ∆E v ) in a constant volume calorimeter qv = ∆E ConstantConstant-Volume Calorimeter Calorimeter OxygenOxygen-bomb Calorimeter Calorimeter Calorimetry Calorimetry Cork is not air tight, not under atmospheric atmospheric pressure (1atm) ∆E p = q p + w ∆E p = q p − p∆V ∆E p + p∆V = q p = ...
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This document was uploaded on 02/26/2014.

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