= + = 0 ∆Euniverse ∆Esystem ∆Esurroundings 6.4 Determining the work due to a change in volume at constant pressure (PV work) (258): w = -P∆V Relating the enthalpy change to the internal energy change at constant pressure (258): 6.5 = + P∆V ∆H ∆E 6.6 Identifying the enthalpy change with the heat absorbed or released at constant pressure (259): qP = + P∆V = ∆E ∆H 6.7 Calculating the heat absorbed or released when a substance undergoes a temperature change or a reaction occurs (261): q = c × mass × ∆T 6.8 Calculating the overall enthalpy change of a reaction (Hess’s law) (268): = + + ··· + ∆Hoverall ∆H1 ∆H2 ∆Hn 6.9 Calculating the standard enthalpy of reaction (271): = - m∆H°f n∆H°f ∆H°rxn (products) (reactants) BRIEF SOLUTIONS TO FOLLOW-UP PROBLEMS 6.1A = q + w ∆E = [13.5 kJ + (-1.8 kJ)] × = 1.17 × 10^4 J 6.1B = q + w ∆E ) + (15.0 Btu × ) =(-26.0 kcal × = -93 kJ (atm) = 732 torr × = 0.963 atm 6.2A P w (atm·L) = -P∆V = -0.963 atm × (2.35 L - 5.68 L) = 3.21 atm·L atm·L w (J) = 3.21 × = 325 J 6.2B w (atm·L) = -P∆V = -5.5 atm × (16.3 L - 10.5 L) = -32 atm·L atm·L w (J) = -32 × = -3.2×10^3 J 6.3A The reaction is exothermic. 6.3B The process is endothermic.
∆Euniverse = ∆Esystem + ∆Esurroundings = 0 6.4 Determining the work due to a change in volume at constant pressure (PV work) (258): w = -P∆V 6.5 Relating the enthalpy change to the internal energy change at constant pressure (258): ∆H = ∆E + P∆V 6.6 Identifying the enthalpy change with the heat absorbed or released at constant pressure (259): qP = ∆E + P∆V = ∆H 6.7 Calculating the heat absorbed or released when a substance undergoes a temperature change or a reaction occurs (261): q = c × mass × ∆T 6.8 Calculating the overall enthalpy change of a reaction (Hess’s law) (268): ∆Hoverall = ∆H1 + ∆H2 + ··· + ∆Hn 6.9 Calculating the standard enthalpy of reaction (271): ∆H°rxn = m∆H°f (products) - n∆H°f (reactants) BRIEF SOLUTIONS TO FOLLOW-UP PROBLEMS 6.1A ∆E = q + w = [13.5 kJ + (-1.8 kJ)] × = 1.17 × 10^4 J 6.1B ∆E = q + w =(-26.0 kcal × ) + (15.0 Btu × ) = -93 kJ 6.2A P (atm) = 732 torr × = 0.963 atm w (atm·L) = -P∆V = -0.963 atm × (2.35 L - 5.68 L) = 3.21 atm·L w (J) = 3.21 atm·L × = 325 J 6.2B w (atm·L) = -P∆V = -5.5 atm × (16.3 L - 10.5 L) = -32 atm·L w (J) = -32 atm·L × = -3.2×10^3 J 6.3A The reaction is exothermic. 6.3B The process is endothermic.
ch6-pg-273-285.docx