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5 Thermochemistry Solutions to Exercises Enthalpy (sections 5.3 and 5.4) 5.33 (a) Change in enthalpy is usually easier to measure than change in internal energy because, at constant pressure, = The heat flow associated with a process at constant pressure can easily be measured as a change in temperature. Measuring requires a means to measure both q and W. (b) H describes the enthalpy of a system at a certain set of conditions; the value of H depends only on these conditions. q describes energy transferred as heat, an energy change which, in the general case, does depend on how the change occurs. We can equate change in enthalpy, with heat, only for the specific conditions of constant pressure and exclusively P-V work. (c) If AH is positive, the enthalpy of the system increases and the process is endothermic. 5.34 (a) When a process occurs under constant external pressure and only P-V work occurs, the enthalpy change (AH) equals the amount of heat transferred. AH = (b) = If the system releases heat, q and are negative and the enthalpy of the system decreases. (c) If = 0, qₚ=0 = and = W. 5.35 At constant pressure, = AH - PAV. In order to calculate more information about the conditions of the reaction must be known. For an ideal gas at constant pressure and temperature, PAV = RTAn. The values of either P and or T and An must be known to calculate from AH. 5.36 At constant volume (V = 0), = According to the definition of enthalpy, H = E + PV, so AH = + For an ideal gas at constant temperature and volume, APV = VAP = RTAn. For this reaction, there are 2 mol of gaseous product and 3 mol of gaseous reactants, so An = -1. Thus VAP or A(PV) is negative. Since AH = + A(PV), the negative term means that AH will be smaller or more negative than E. 5.37 Analyze/Plan. q = 824 J = 0.824 kJ (heat is absorbed by the system), W = 0.65 kJ (work is done on the system). Solve. = q + W = 0.824 kJ + 0.65 kJ = 1.47 kJ. = q = 0.824 kJ (at constant pressure). Check. The reaction is endothermic. 5.38 The gas is the system. If 0.49 kJ of heat is added, q = +0.49 kJ. Work done by the system decreases the overall energy of the system, so W = -214] = -0.214 kJ = q + W = 0.49 kJ 0.214 kJ = 0.276 kJ. = q = 0.49 kJ (at constant pressure). 5.39 (a) + 3H₂O(g) + = -1235 kJ 114