A 4.400 g sample of propane (C 3 H 8 ), as gas used in portable gas cookers, was

Question

A 4.400 g sample of propane (C3H8), as gas used in portable gas cookers, was placed in a calorimeter where it was burned in the presence of excess oxygen, O2.

C3H8(g) + 5 O2(g) ? 3 CO2(g) + 4 H2O(l)

The water bath surrounding the calorimeter contained 1.000 kg of water at an initial temperature of 25.000°C. The reaction raised the temperature of the water surrounding the calorimeter to 27.828°C.

a) How many joules (J) of heat were absorbed by the water surrounding the calorimeter?

The specific heat of water is 4.184 J g-1K-1

The molar mass of propane is 44.10 g mol-1

b) If the heat capacity of the calorimeter is 93.1 kJ K-1, how much heat was absorbed by the calorimeter?

c) What is the total heat (in kJ) released by the combustion of the 4.400 g of propane?

d) What is the enthalpy of reaction (in kJ mol-1) of propane with oxygen?

Explanation / Answer

Ans. #a. Heat absorbed by water, q1 = m s dT

            Where, m = mass of water in grams

                        s = specific heat of water

                        dT = Increase in temperature

# Putting the values in above expression –

            q1 = 1000.0 g x 4.184 J g-1 K-1 x [(27.828 + 273.15)K – (25.000 + 273.15)K ]

            q1 = 1000.0 g x 4.184 J g-1 K-1 x 2.828 K

            Hence, q1 = 11832.352 J = 11.832 kJ

#b. Heat absorbed by Calorimeter, q2 = C x dT

            Where, C = heat capacity of calorimeter

Now, q2 = 93.1 kJ K-1 x 2.828 K = 263.287 kJ

#c. Total heat released from combustion of 4.400 g propane must be equal to the total amount of heat absorbed by water and calorimeter.

So,

            Heat released from combustion of propane, Q = - (q1 + q2)

                                                            = - (11.832 kJ + 263.287 kJ)

                                                            = -275.119 kJ

Note: The –ve sign of Q indicates that heat is being released during combustion.

#d. Moles of propane combusted = Mass/ MW = 4.400 g / 44.10 g mol-1

                                                            = 0.09977 mol

Now,

            Molar enthalpy of reaction = Q / Moles of propane

                                                            = (-275.119 kJ) / 0.09977 mol

                                                            = -2650.77 kJ mol-1

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