THERMODYNAMICS PRACTICE PROBLEMS 1. Tungsten is obtained ly by the on of WOs wit
ID: 637922 • Letter: T
Question
THERMODYNAMICS PRACTICE PROBLEMS 1. Tungsten is obtained ly by the on of WOs with hydrogen per the equation below The following data related to this reaction are available 2 840.3 763.4 241.8 228.4 AGP kJ/mol (a) Calculate ??, ?G and AS at 25°C for the reaction indicated by the equation above (b) What is the valuc of the equilibrium constant for the system represented above? (c) Find the temperature at which the reaction mixture is in equilibrium (AG-0) at 1 atmosphere. 2. The cnthalpy change for the combustion of butyric acid at 25?' ?/r ?s-2,183.5 kilojoules per mole. I Se J mol K Afte kJ mol 0,00 Substance 5.69 13.6 130.6 69.91 93.5 Hz 0.00 285.85 ILO 41.8 (a) From the above data, calculate the standard heat of formation, AHP. for butyric acid. (b)Calculate the change in entropy for the reaction given above (c) Calculate the change in free energy for the reaction given above. (d) Use the values of ??, ?S and ?G to describe the reaction. 3. Glucose, CoHizO%, is readily metabolized in the body. a. Write a balanced equation for the metabolism of CoHizOs to CO: and H-O. b. Calculate ??,metabolism for glucose. Given: Substance AGP kJ mol 394.4 237.2 CO2 0.00 917 c. IfAF for this process is -2801.3 kJ, calculate AS at 25 °c d. One step in the utilization of encrgy in cells is the synthesis of ATP from ADP and HaPO AG-30.5 kJ/mol i. Calculate the equilibrium constant, K, for the formation of ATP at 25 °C ?Calculate the number of moles of ATP" formed by the metabolism of 1.0 g of glucose. (hint: determine how much free energy is produced by the metabolism of 1,0 g of glucose) . The enthalpy of combustion of liquid octane, Caffu) to gaseous products, is -5090 kJ/mol-1. Answer the questions elow, assuming a temperature of 100 ?. a. write a balanced cquation for the complete combustion of liquid octane Calla() b. Dctcrmine the molar cnthalpy of formation, ? le, for liquid octane. Callar) (see problem #2 for All values) C. If?03° for the combustion is-5230 kJ/mol of octane, calculate the value of AS. Comment on the sign of AS? relative to the equation written above.
Explanation / Answer
Ans 1
Part a
Enthalpy change for the reaction = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants
H = 3*Hf(H2O) + Hf(W) - 3*Hf(H2) - Hf(WO3)
= 3*(-241.8) + 0 - 3*0 - (-840.3)
= 114.9 kJ/mol
Standard free energy change for the reaction = sum of Standard free energy of formation of products - sum of Standard free energy of formation of reactants
G = 3*Gf(H2O) + Gf(W) - 3*Gf(H2) - Gf(WO3)
= 3*(-228.4) + 0 - 3*0 - (-763.4)
= 78.2 kJ/mol
G = H - TS
S = (H - G) / T
= (114.9 - 78.2) /298
= 0.123 kJ/mol-K x 1000J/kJ
= 123 J/mol·K
Part b
G = - RT ln K
78.2 *1000 J/mol = - 8.314 J/mol·K x 298K x ln K
ln K = - 31.563
K = 1.96 x 10^-14
Part C
At equilibrium
G = H - TS
0 = 114.9 - T*0.123
T = 934.14 K
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