C102Set2SolnRevS15 - Page 1 SOLUTION TO PROBLEM SET 2(1a C2H5OH(l 3 O2(g-> 2 CO2(g 3 H2O(l Balanced combustion reaction for one mole of ethanol(1b

C102Set2SolnRevS15 - Page 1 SOLUTION TO PROBLEM SET 2(1a...

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Page 1 SOLUTION TO PROBLEM SET # 2 (1a) C 2 H 5 OH(l) + 3 O 2 (g) ------> 2 CO 2 (g) + 3 H 2 O(l) - Balanced combustion reaction for one mole of ethanol. (1b) q system or total = 0 = q rxn + q cal (insulated). Also, in a bomb calorimeter , all of the q's are measured at constant volume, i.e., q v . Thus: q rxn = - q cal = - C cal T ; where T = T f - T i , it is the same for all of the contents, since they are in equilibrium. T = 26.941ºC - 25.000 o C = +1.941ºC. q rxn = - 15.259 kJ ºC (+1.941 ºC)} = - 29.618 kJ for 1.000 g of ethanol. q rxn (per mole) = - 29.618 kJ 1.000 g ethanol 46.08 g ethanol 1 mole ethanol = - 1364.8 kJ/mole ethanol. q v,rxn (per mole) = E rxn (per mole) = -1364.8 kJ/mole ethanol . Note: q v = E since E = q + w = q - P ext V = q v ; because at constant volume, V, and hence w is zero . (1c) As discussed in class, q v,rxn = E rxn measured in (b) is at the initial temperature of the system . Thus, we have: H rxn,298 E rxn,298 + (PV) gases = E rxn,298 + n g RT ; where T is equal to 298 K ( T initial for E rxn = q v,rxn ). n g = (2 - 3) mol = - 1 moles. R = 8.314 x 10 - 3 kJ/mol-K. Plugging in: H rxn,298 = - 1364.8 kJ + ( - 1 moles)(8.314 x 10 - 3 kJ/mol-K)(298 K) H rxn,298 = -1367.28 kJ/(mole ethanol) . (1d) Assuming that H rxn,298 measured in (c) is equal to H o rxn,298 (i.e., assuming standard conditions), then we can say (according to Hess): H rxn,298 = - 1367.28 kJ/mol - 1367.28 kJ/mol = (2) H o f,298 (CO 2 ,g) + (3) H o f,298 (H 2 O,l) - (1)• H o f,298 (C 2 H 5 OH,l). Solving (suppressing the 298 K subscript throughout): H o f (C 2 H 5 OH,l) = ( ) (2) H o f (CO 2 ,g) + (3) H o f (H 2 O,l) - ( - 1367.28) kJ mole ethanol . Thus, plugging in: H o f (C 2 H 5 OH,l) = ((2)( - 393.5)+ (3)( - 286) - ( - 1367.28)) kJ/mole ethanol. H o f (C 2 H 5 OH,l) = -277.72 kJ/mol ethanol.
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Page 2 2. Rxn .: C 2 H 4 (g) + 3 O 2 (g) ------> 2 CO 2 (g) + 2 H 2 O(l) Given : reversible, constant pressure of 1 atm and a constant temperature of 25 o C or 298 K and H o rxn,298 = - 1410.8 kJ. Thus, from the given conditions: H rxn = H o rxn,298 = q p,rxn . So, for the reaction, as written, H rxn = - 1410.8 kJ and q rxn = -1410.8 kJ . From the definition of H, we have : E rxn = H rxn - (PV) rxn H rxn - [(PV) product,gases - (PV) reactant,gases ] ; where we have neglected the “PV” term (i.e., volume) for H 2 O(l). Using the ideal gas law, PV = n g RT, for each of the gases, realizing that temperature (T) is constant - we have: E rxn H rxn + [(n g,prod - n g,reac )RT] = H rxn - n g RT ) (where the “ ” applies only to n g not to “RT”) Since we are at standard conditions (1 atm) and also, since the temperature is 298 K, we have: E rxn = 298 and H rxn = 298 ) - by definition. Thus, H o 298 = -1410.8 kJ . Substituting, we have: 298 = 298 - n g RT = ( - 1410.8 kJ) - (2 mol - 4 mol) 8.314 x 10 - 3 kJ mol K (298 K) 298 = - 1410.8 kJ + 4.955 kJ = - 1405.8 kJ => 298 = -1405.8 kJ . We can get work (w) from the first law of thermodynamics: E = q + w . Solving for w: w = E - q = - 1405.8 kJ - ( - 1410.8 kJ) = +5 kJ => w = +5 kJ . Thus, work is done on the system by the surroundings .
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