15) CH,OH(s)+7 0-(g)- 6 CO(g)+3 H,00 When a 2.000-gram sample of pure phenol, C&

Question

15) CH,OH(s)+7 0-(g)- 6 CO(g)+3 H,00 When a 2.000-gram sample of pure phenol, C&H; OH(s). is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow. Standard Heat of Formation, AHP, Absolute Entropy, Ss°, Substance C(graphite CO (g) H2(g) H20() O2(g) C&H; OH(s) at 25°C (kJ/mol) -393.5 -285.85 5.69 0.00 0.00 0.00 213.6 130.6 69.91 205.0 144.0 2 (a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C (b) Calculate the standard heat of formation, AH?, of phenol in kilojoules per mole at 25°C. (c) Calculate the value of the standard free-energy change, AGo, for the combustion of phenol at 25°C. If the volume of the combustion container is 10.0 liters, calculate the final pressure in the container when the temperature is changed to 110.°C. (Assume no oxygen remains unreacted and that all products are gaseous.) (d)

Explanation / Answer

C6H5OH (s) + 7 O2(g) ------> 6 CO2 (g) + 3 H2O (l)

number of moles of Phenol = 2/94 = 0.0213 moles

0.0213 moles of Phenol produces 64.98 kJ heat

1 mole of phenol produces 64.98/0.0213*1 = 3050.704 kJ

so 1 mole of phenol on combustion will produce 3050.704 kJ of heat

so we can write the equation as,

C6H5OH (s) + 7 O2(g) ------> 6 CO2 (g) + 3 H2O (l) ; dH rxn = -3050.704 kJ/mol

dH rxn = dH products - dH reactants

-3050.704 = (6*(-393.5)+3*(-285.85)) - (x-0)

x = dH formation of Phenol = -167.846 kJ/mol

C6H5OH (s) + 7 O2(g) ------> 6 CO2 (g) + 3 H2O (l)

dS rxn = dS products - dS reactants

dS rxn = ((6*213.6)+(3*69.91))-(144+205)

dS rxn = 1142.33 J/mol.K

dG = dH - TdS

dG = (-3050.7 * 10^3) - (298*1142.33)

dG = -3391 kJ/mol

let us consider P1V1/n1T1 = P2V2/n2T2

here volume of the container V1 = v2 = 10 L

1 mole of Phenol requires 7 moles of O2

0.0213 moles of Phenol requires 7 * 0.0213 moles of O2

number of moles of O2 = 0.1491 moles

number of moles of gas initially = 0.1491 moles of O2 =n1

1 mole of phenol produces 6 mole of CO2

0.0213 moles of phenol produces 6*0.0213 moles of CO2 = 0.1278 moles

number of moles of gas finally = 0.1278 moles of CO2 = n2

initial temperature = T1 = 25 0C = 298 K

final temperature = T2 = 110 0C = 110 + 273 = 383 K

then using the formula,

1*10/(0.1491*298) = P2*10/(0.1278*383)

P2 = final pressure in the container = 1.1016 atm

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