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Transcript
First Law of Thermodynamics (Law of Conservation of Energy) The combined amount of matter and energy in the universe is constant. Potential energy Kinetic energy ΔE = q + W What are the symbols? Internal Energy: ΔE Heat transferred (into or out of system): q Work done (on or by the system): w Enthalpy = heat gained or lost by system: ΔH ΔH < 0: Exothermic process Mullis 1 q and ΔH: Signs Sign is determined by the experience of the system: Heat out Heat in System (The reaction occurs in here!) System (The reaction occurs in here!) ΔH > 0 q>0 Both q & Δ H are positive. Endothermic Rxn ΔH < 0 q<0 Both q & Δ H are negative. Exothermic Rxn Mullis 2 Relationships: ΔH, ΔE, q and w ΔE = q + w ΔE = ΔH - P ΔV (P ΔV is the amount of work done by expanding gases, but in most reactions, there is a very small volume change, so ΔE ~ ΔH in many cases.) q , ΔH q , ΔH w w ΔE ΔE Sign + + + - Condition Heat transferred to system Heat transferred from system Work done by surroundings Work done by system q > 0 and w > 0 q < 0 and w < 0 Mullis 3 Internal Energy Internal energy = ΔH – PΔV = ΔH – ΔnRT If number of moles of gas change in the reaction, include the # moles term. The combustion of C3H8(g) to produce gaseous products has ΔHr = -2044.5 kJ-mol-1 at 298 K. Find the change in internal energy for this reaction. C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) Δn = 7-6 = 1 ΔnRT = 1 mol(8.314 J mol-1K-1)(298K) = 2478J Int. Energy = -2044.5 kJ -2.478 kJ = -2047 kJ Mullis 4 Calorimetry Calorimeter = Measures temp change in a process “Bomb” calorimeter = constant volume Under constant pressure, heat transferred = enthalpy change ( q = ΔE ) Heat capacity = amount of heat to raise temp by 1K (or 1º C) Specific heat = heat capacity for 1 g of a substance (Symbol for specific heat is usually C) Amount of heat absorbed by a substance calculated using mass, specific heat and temperature change: q = ΔE = mc Δ T Mullis 5 Heat Measurements Using Calorimeter 50.0 mL of 0.400 M CuSO4 at 23.35 ºC is mixed with 50.0 mL of 0.600 M NaOH at 23.35 ºC in a coffee-cup calorimeter with heat capacity of 24.0 J/ ºC. After reaction, the temp is 25.23 ºC. The density of the final solution is 1.02 g/mL. Calculate the amount of heat evolved. Specific heat of water is 4.184J/g ºC . q = ΔE = mc Δ T Mass = (50 mL+50 mL)(1.02g/mL) = 102 g Δ T = 25.23-23.35 = 1.88 ºC Heat absorbed by solution = (102g)(4.18J)(1.88 ºC) = 801J g ºC Add this to heat absorbed by calorimeter = (24.0J)(1.88 ºC) = 45.1J g ºC Total heat liberated by this reaction = 846 J Mullis 6 Bomb Calorimeter 1.00 g ethanol, C2H5OH, is burned in a bomb calorimeter with heat capacity of 2.71 kJ/g ºC. The temp of 3000 g H2O rose from 24.28 to 26.22 ºC. Determine the ΔE for the reaction in kJ/g of ethanol and then in kJ/mol ethanol. Specific heat of water is 4.184J/g ºC . q = ΔE = mc Δ T Δ T = 26.22 - 24.28- = 1.94 ºC Heat to warm H2O = (3000g)(4.18J)(1.94 ºC) = 24.3 x 103J = 24.3 kJ g ºC Add this to heat to warm calorimeter = (2.71kJ)(1.94 ºC) = 5.26 kJ g ºC Total heat absorbed by calorimeter and water = 29.6 kJ = 29.6 x 103J Since this is combustion, heat is liberated so change sign for heat of reaction: - 29.6kJ/g ethanol -29.6kJ | 46.1 g = -1360 kJ ethanol g | mol mol Mullis 7 Hess’s Law Law of heat summation The enthalpy change for a reaction is the same whether it occurs by one step or by any series of steps. A state function such as enthalpy does not depend on the steps State functions are analogous to your bank account: Many combinations of deposits and withdrawals result in what you watch: The account balance. ΔHºrxn = ΔHºa + ΔHºb + ΔHºc +… Application is an accounting/algebra exercise to get resulting reaction ΔHºrxn by adding reactions of known ΔH. Mullis 8 Hess’s Law Write resulting equation and arrange the step equations to get products in step reactions on same side of products in the final reaction. If reaction is reversed, change sign of ΔH. Multiply step equations as needed to get the resulting equation. Fractions are allowed. Recall that coefficients correspond to moles. Mullis 9 Hess’s Law Example C2H5OH + 3O2 2CO2 + 3H2O ΔH = -1367 kJ/mol C2H4 + 3O2 2 CO2 + 2H2O ΔH = -1411 kJ/mol Find ΔH for C2H4 + H2O C2H5OH 2CO2 + 3H2O C2H5OH + 3O2 ΔH = 1367 kJ/mol C2H4 + 3O2 2 CO2 + 2H2O ΔH = -1411 kJ/mol ___________________________________________ C2H4 + H2O C2H5OH ΔH = -44 kJ/mol Mullis 10