CHM1045 Lecture 18 Outline of Last Lecture:I. EnergyII. 2 Types of Processes:III. Thermodynamics: IV. State Functions:V. First law of thermodynamics VI. Work Done By the System On the SurroundingsVII. Enthalpy and the First Law of ThermodynamicsVIII. Enthalpy:IX. Thermochemical EquationsX. A Comparison of ∧H and ∧UOutline of Current Lecture:I. HeatII. Constant-Volume CalorimetryIII. Constant-Pressure CalorimetryCurrent Lecture:Example 1:Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25°C:∧ n= 2-3∧ n=-1 ∧ U=∧ H-RT∧ n = -566.0kj/mol-(8.314jk*mol)(1kj/1000J)(298K)(-1) = -563.5 kl/molHeat- specific heat(s): a substance is the amount of heat (q) required to raise the temperatureof one gram of the substance by one degree Celsius.- heat capacity (C): a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.Heat ( q ) absorbed or released:q = m x s x Dtq = C x ∧tDt = tfinal - tinitialExample 2:A 466-g sample of water is heated from 8.50°C to 74.60°C. Calculate the amount of heat absorbed (in kilojoules) by the water.Q=ms∧ t = (466g)(4.184j/g*C)(74.60C-8.50C) = 1.29*10^5*1Kj/1000j= 129jkConstant-Volume Calorimetryqsys = qwater + qbomb + qrxnqsys = 0qrxn = - (qwater + qbomb)qwater = m x s x ∧tqbomb = Cbomb x ∧tReaction at Constant V∧H ≠ qrxn∧H ~ qrxnExample 3:A quantity of 1.435 g of naphthalene (C10H8), a pungent-smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28°C to 25.95°C. If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.Qcal-Ccal ∧ t = (10.17kj/c)(25.95C-20.28C)= 57.666kJ 57.666kJ * 128.2g C10H8 = -5.51*10^3 kJ/mol 1.435C10H8 1 mol C10H8Constant-Pressure Calorimetryqsys = qwater + qcal + qrxnqsys = 0qrxn = - (qwater + qcal)qwater = m x s x ∧tqcal = Ccal x ∧tReaction at Constant P∧H = qrxnExample 4:A lead (Pb) pellet having a mass of 26.47 g at 89.98°C was placed in a constant-pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50°C to 23.17°C. What is the specific heat of the lead pellet?Qh20=(100.0g)(4.184kJ*C)(23.17-22.5)= 280.3Jqpb= ms∧ t-280.3J = (26.47)(s)(23.17C-89.98)s= .158
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