One mole of propane, C3H8 is placed in an adiabatic (isolated), constant pressur

Question

One mole of propane, C3H8 is placed in an adiabatic (isolated), constant pressure combustion chamber at 25C and allowed to react with a stoichiometric amount of oxygen and reacts to form H2O and CO2

1 C3H8 + 5 O2 -----> 4 H2O (g) + 3 CO2 (g)

Delta H C3H8 = -103.85 kJ/mole, Delta H CO2= -393.51 kJ/mole, Delta H H2O= -241.82 kJ/mole.

What is the enthalpy change for the reaction?

Using the information above, assum complete combustion of the one mole of propane, gas ideal behavior, and all heat generated from combustion will be gained by CO2 and H2O. The specific heat values are as follows: C CO2= 56.91 J/mole*K and C H20= 58.56 J/mole*K. What is the final temperature?

Using the information above and a volume of 150 kL. What are the mole fractions and the partial pressures of H2O and of CO2?

Explanation / Answer

enthalpy change = (deltaH CO2 + deltaH H2O ) - (deltaH C3H8)

enthalpy change = -241.82 - 393.51 + 103.85 = -531.48KJ/mole

when 1 mole of C3H8 completely reacts , mass of H2O formed = 72g(4moles) , mass of CO2 = 132g (3moles)

now , equating enthalpy , -531.48 = (4*58.56 * (25-x)) + (3 * 56.91 * (25-x))

so (25-x) = -531.48 / 404.97 , 25-x = -1.312 , x = 26.312 degrees

total no.of moles = 7 ,

mole fraction of H2O = 4/7 = 0.5714

mole fraction of CO2 = 3/7 = 0.4286

total volume is in the ratio 4: 3 , total volume = 150Kl , volume of H2O = 85.71 Kl , volume of CO2 = 64.29Kl

we know that Vgas / Vtotal = Pgas / Ptotal , so for H2O , 85.71/150=Ph2o / Ptotal =0.5714

similarly Pco2 / ptotal = 0.4286 , Ptotal = 1atm

so Ph20 = 0.5714atm , Pco2 = 0.4286atm

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