First Law of Thermodynamics (Law of Conservation of Energy)

1. 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

2. 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 q and ΔH: Signs Sign is determined by the experience of the system: Heat out Heat in Mullis

3. 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.) Mullis

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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 + 3O2ΔH = 1367 kJ/mol C2H4 + 3O2  2 CO2 + 2H2O ΔH = -1411 kJ/mol ___________________________________________ C2H4 + H2O  C2H5OH ΔH = -44 kJ/mol Mullis