Lecture 7: Calorimetry

5.60 Spring 2008 Lecture #7 page 1 Calorimetry The objective is to measure isothermal Reactants (T1) ΔHrx (T1 ) = constant p (for solutions) Constant pressure • Products (T1) CALORIMETER reaction thermometer I) ΔHI React. (T1) + Cal. (T1) II) ΔHII Prod. (T2) + Cal. (T2) Adiabatic Constant p adiabatic = constant p Prod. (T2) + Cal. (T2) Prod. (T1) + Cal. (T1) = constant p ______________________________________________________ ΔHrx (T1 ) React. (T1) + Cal. (T1) = constant p Prod. (T1) + Cal. (T1) ΔHrx (T1 ) = ΔHI + ΔHII (I) Purpose is to measure (T2 - T1) Adiabatic, const. p ⇒ qp = 0 ΔH I = 0 ⇒ (II) Purpose is to measure heat qp needed to take prod. + cal. from T2 back to T1. T2 qp = ∫ C p ( Prod . + Cal . ) dT = ΔHII T1 ∴ T2 T2 1 1 ΔHrx (T1 ) = − ∫T C p ( Prod . + Cal . )dT ≈ − ∫T C pcal dT = −C pcal ΔT 5.60 Spring 2008 • Lecture #7 Constant volume page 2 (when gases involved) CALORIMETER reaction Adiabatic Constant V thermometer I) ΔU I React. (T1) + Cal. (T1) II) ΔUII Prod. (T2) + Cal. (T2) adiabatic = constant V = constant V Prod. (T2) + Cal. (T2) Prod. (T1) + Cal. (T1) ______________________________________________________ ΔUrx (T1 ) React. (T1) + Cal. (T1) = constant V Prod. (T1) + Cal. (T1) ΔUrx (T1 ) = ΔUI + ΔUII (I) Purpose is to measure (T2 - T1) Adiabatic, const. V ⇒ qV = 0 ΔUI = 0 ⇒ (II) Purpose is to measure heat qV needed to take prod. + cal. from T2 back to T1. T2 qV = ∫ CV ( Prod . + Cal . ) dT = ΔUII T1 ∴ T2 T2 1 1 ΔUrx (T1 ) = − ∫T CV ( Prod . + Cal . )dT ≈ − ∫T CVcal dT = −CVcal ΔT Now use H = U + pV or ΔH = ΔU + Δ ( pV Assume only significant contribution to Δ ( pV ) ) is from gases. 5.60 Spring 2008 Lecture #7 Δ ( pV ) = R Δ (nT Ideal gas ⇒ Isothermal T =T1 ⇒ ∴ page 3 ) Δ ( pV ) = RT1 Δngas ΔHrx (T1 ) = ΔUrx (T1 ) + RT1 Δngas T2 ΔHrx (T1 ) = − ∫ CV ( Prod . + Cal . ) dT + RT1 Δngas ≈ −CVcal ΔT + RT1 Δngas T1 Difference between ΔU and ΔH small but measurable e.g. 4 HCl(g) + O2(g) = 2 H2O(l) + 2 Cl2(g) T1 = 298.15 K ΔUrx (T1 ) = -195.0 kJ Δngas = -3 moles ΔHrx (T1 ) = -195.0 kJ + (-3 mol)( 298.15 K)(8.314x10-3 kJ/K-mol) = -202.43 kJ Now let’s imagine really running this reaction in a constant-V calorimeter with Cv = 10 kJ/K Calorimeter thermal mass >> product thermal mass Heat goes to changing calorimeter T Often no need to know product Cp or CV value 5.60 Spring 2008 Lecture #7 page 4 Bond energies: An approximate method for estimating ΔHf° Really bond enthalpies, but usually Δ(pV) << difference 1) 2) e.g. Measure bond energies for known compounds Use them to estimate ΔHf° for unknown compounds CH4(g) = C(graphite) + 2 H2(g) ΔHI = −ΔHf°,CH C(graphite) = C(g) ΔHC ( atomization ) 4 2ΔHH ( atomization ) 2 H2(g) = 4 H(g) 2 ____________________________________________________________ ΔH CH4(g) = C(g) + 4 H(g) H | ΔH = 4BC-H H- C -H | BC-H = 416.2 kJ H 4 BC-H = −ΔHf°,CH + ΔHC (atom .) + 2ΔHH 4 ⇒ BC −H = 416.2 kJ 2 ( atom .) H H C2H6(g) = C(g) + 6 H(g) | | | | H- C - C -H H H ΔH = BC-C + 6BC-H = −ΔHf°,C H + 2ΔHC (atom .) + 3ΔHH (atom .) ⇒ BC −C = 342 kJ 2 6 2 Now estimate ΔHf° for n-pentane, C5H12 CH3-CH2- CH2- CH2-CH3 5 C(graphite) = 5 C(g) 5ΔHC (atom .) 6 H2(g) = 12 H(g) 6ΔHH 5 C(g) + 12 H(g) = C5H12(g) 2 ( atom .) ΔH ≈ −(4Bc −c + 12B C −H ____________________________________________________________ 5 C(graphite) + 6 H2(g) = C5H12(g) ΔHf°,C H 5 12 ) 5.60 Spring 2008 Lecture #7 page 5 ΔHfo,C 5H12 ≈ −(4Bc −c + 12B C −H ) + 5ΔHC (atom .) + 6ΔHH2( atom .) ~ -152.6 kJ (estimated) Actual ΔHf°,C H (n-pentane ) = -146.4 kJ 5 12 CH3 But | is also C5H12 with 4 C-C bonds, 12 C-H bonds CH3 - C -CH3 | CH3 ⇒ ΔHf°,C H ~ -152.6 kJ (estimated using bond energies) 5 12 Actual ΔHf°,C H (neopentane ) = -166.1 kJ 5 12