Substance C (s) co23) 12 H20 0) 023) 5.69 213.6 1306 6991 205.0 2263 3935 285.85
ID: 574780 • Letter: S
Question
Substance C (s) co23) 12 H20 0) 023) 5.69 213.6 1306 6991 205.0 2263 3935 285.85 1. When 2.50 grams of butyric acid, C3H7COOH, CO2 (g) and H20() are formed with the evolution of 600M acid, C3HyCoOH, is completely combusted at 25 C and 1.00 a of combustion (a) Calculate the molar heat of combustion for butyric acid. (b Write a balanced thermochemical e (c) Calculate the standard (d) If 5.00 g of butyr combustion for butyric acid. equation for the molar heat tree energy of combustion, AGf, for butyric acid at 25 C. ric acid is combusted in a bomb calorimeter, what will be the heat 30.0 capacity of the bomb if the temperature of the bomb goes from 25.0 If 5.00 g of butyric acid is combusted in the part (d) bomb calorimeter surrounded by 1.912 kg of water at 25'C, what will be the final temperafure that is (e) the system? The specific heat of water is 4.184 J/gC. butyric acid. butyric acid. at 25'C ta at the top, calculate the standard heat of fom ration, 1rf, for (G) Write a balanced thermochemical equation for the molar heat of formation e the standard entropy change, AS's, for the formation of butyric acid (i) Calculate the standard free energy of formation AG.f, for butyric acid at 25 G) Would the molar heat of formation for H20 (3) be (more, less, or the same) when Explain. Calculate the volume of air at 25'C and 750. mmHg that is needed to combust 20.0 grams of butyric acid. Air is 200 percent by volume oxygen. compared to the molar heat of formation for H207 (k) Predict how the enthalpy, entropy, free energy, and temperature change in the system during this process. Explain the basis for each of your predictions. 2. (a) When liquid water is introduced into an evacuated vessel at 25'C, some of the water vaporizes. the temperature of the system decreases. Preidct how the enthalpy, entropy, and free energy change in the system during this process. Explain the basis for each of your predictions (b) When a large amount of ammonium chloride is added to water at 25'C, some of it dissolves and If the temperature of the aqueous ammonium chloride system in part (b) were to be increased to 30C predict how the solubility of the ammonium chloride would be affected. Explain the basis for each of your predictions. (c)Explanation / Answer
calculate the moles of butyric acid with
moles = grams / molar mass
molar mass of butyric acid 88.11 g/gmol
moles = 2.5 / 88.11 = 0.02837 moles
molar heat combustion is the energy released per mole combusted so
moles = 60 KJ / 0.02837 moles = 2114.64 KJ / mole
B)
Butyric acid will have a molecular formula like
C4H8O2 so the equation will be
C4H8O2 + O2 ==== CO2+ H2O (assuming a complete combustion) we have to balance this , we have 8 hydrogens on the left so we have to multiply by 4 the h2o
C4H8O2 + O2 ==== CO2+ 4 H2O, we have 4 carbon on the left and 1 on the right so multiply by 4 the CO2
C4H8O2 + O2 ==== 4 CO2+ 4 H2O, we have 4 oxygens on the left and 12 oxygens on the right, multiply by 5 the O2
C4H8O2 + 5O2 ==== 4 CO2+ 4 H2O
C) Gibbs, to do this we have to calculate the entroy
entropy of reaction
S rxn = S products - S reactants
S products = n products * Sformation
S reactants = n reactants * Sformation
S products = 4 * (213.6) + 4*(69.91) = 1134.04 J / K mol
S reactants = 226.3 + 5*205 = 1251.3 J/k mole
S rxn = 1134.04 - 1251 = -117.26 J / K mol
Enthalpy of rxn = -2114.64 KJ / mole, the negative sign is because this reaction releases heat we use the same value than we found in part A, combustion is a reaction itself
Gibbs= H - TS = -2114.64 - (-0.11726*298.15)= -2079.678 KJ / mole, the 117 must be divided by 1000 to get KJ
D) For the heat capacity of the calorimeter the equation we have to apply is
q cal = C x (Tf - Ti), c is the heat capacity
5 grams is equal to 0.057 moles of butyric acid
multiply this value by the enthalpy of combustion we got on part A
0.057 * 2114.64 KJ / mole = 120 KJ of heat released, then this is the heat absorbed by the calorimeter so
120 KJ = C x (30 - 25)
120 = C x (5)
C = 24 KJ / C
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