2. When 1.000 mole of butane (C4H10) undergoes complete combustion, 2877 k of he

Question

2. When 1.000 mole of butane (C4H10) undergoes complete combustion, 2877 k of heat are released. In a particular experiment, 12.6 L of butane, measured at 23.6°C and 738 mmHg, are burned in the presence of a stoichiometric amount of oxygen measured at the same temperature and pressure. 13 CH,o(g) +-02(g) 4C02(g) + 5H20( ) a. How many kilojoules of heat energy are released? (MW of C4H10 is 58.11 g mol ) b. How many kilojoules of work is performed in this reaction? Is the work performed by the system or by the surroundings? Assume that this reaction takes place at a constant pressure of 738 mmHg, and that the final volume of the system is determined at 23.6

Explanation / Answer

Ans. Given,

            Volume, V = 12.6 L

            Temperature, T = 23.60C = 296.75 K

            Pressure, P = 738 mmHg = (738 / 760) atm = 0.971 atm

Now,

Using Ideal gas equation:    PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K) = (0C + 273.15) K

Putting the values in equation 1-

            0.971 atm x 12.6 L = n x (0.0821 atm L mol-1K-1) x 296.75 K

            Or, n = 0.502176 mol

Therefore, moles of butane taken = 0.502176 mol

#1. Amount of energy released = Moles of butane taken x Molar enthalpy of combustion

                                                = 0.502176 mol x (2877.0 kJ/ mol)

                                                = 1444.76 kJ

#2. Given, O2 is taken in stoichiometric amount, i.e. just sufficient amount of O2 for complete combustion.

1 mol butane requires 6.5 mol O2.

So, total moles of O2 taken = 6.5 x moles of Butane

                                    = 6.5 x (0.502176 mol)

                                    = 3.264144 mol

Total initial moles of reactant gases = 0.502176 mol (butane) + 3.264144 mol (O2)

                                    = 3.76632 mol

# Following stoichiometry of balance reactions, 1 mol butane produces 4 mol CO2.

So,

            Moles of CO2 produced = 4 x moles of butane

                                                = 4 x 0.502176 mol

                                                = 2.008704 mol

Note that water formed during combustion is in liquid form. The volume of liquid is negligible compared to gases.

# Net change in number of moles of gases =

Total moles of products formed – Total moles of reactants taken

= 3.76632 mol - 2.008704 mol

= -1.757616 mol

The negative sign of moles indicates that there is loss of moles during combustion.

# Equal moles of gases occupy equal volume.

So, loss is number of moles of gases during the process is equal to the reduction in equivalent volume.

Again using ideal gas equation, the reduction in volume is equal to the volume occupied by 1.757616 mol gas-

            Putting the values in equation 1-

            0.971 atm x V = 1.757616 x (0.0821 atm L mol-1K-1) x 296.75 K

            Hence, V = 42.821 L

Therefore, change (reduction) in volume after combustion, dV = 42.821 L

# Wrok done, W = p (dV)              - at constant pressure

                        = 0.971 atm x 42.821 L

                        = 41.579191 atm L                                       ; [1 atm L = 101.33 J]

                        = 41.579191 x (101.33 J)

                        = 4213.22 J

                        = 4.212 kJ

Work is done on the system by the surrounding because there is net reduction in volume that would cause shrinkage of the reaction vessel.

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