Question

The gas butane, C₄H₁₀(g), can be used in welding. When butane is burned in oxygen, the reaction is:

2 C₄H₁₀(g) + 13 O₂(g) ---> 8 CO₂(g) + 10 H₂O(g)

(a) Using the following data, calculate ΔH° for this reaction.

ΔH°_f kJ mol^-1: C₄H₁₀(g) = -125.7; CO₂(g) = -393.5 ; H₂O(g) = -241.8

ΔH° = ______

(b) Calculate the total heat capacity of 8 mol of CO₂(g) and 10 mol of H₂O(g), using C_CO2(g) = 37.1 J K^-1 mol^-1 and C_H2O(g) = 33.6 J K^-1 mol^-1.

C= ______

(c) When this reaction is carried out in an open flame, almost all the heat produced in part (a) goes to raise the temperature of the products. Assuming that the reactants are at 25°C, calculate the maximum flame temperature that is attainable in an open flame burning butane in oxygen. The actual flame temperature would be lower than this because heat is lost to the surroundings.

Maximum temp in degrees C = _____

Answer 1

ANSWER:

The given reaction is :

C₃H₈(g) + 5 O₂(g) 3 CO₂(g) + 4 H₂O(g)

(a) Given,

ΔH°_f {C₃H₈(g)} = -103.8 kJ mol^-1 ;

ΔH°_f {CO₂(g)} = -393.5 kJ mol^-1 ;

ΔH°_f {H₂O(g)} = -241.8 kJ mol^-1

Total ΔH° for this reaction is:

ΔH°_reaction = ΔH°_product - ΔH°_reactant

= 3 x ΔH°_f {CO₂(g)} + 4 x ΔH°_f {H₂O(g)} - [1 x ΔH°_f {C₃H₈(g)} + 5 x ΔH°_f {O₂(g)}]

= 3 x {-393.5 kJ mol^-1} + 4 x {-241.8 kJ mol^-1} - [1 x {-103.8 kJ mol^-1} + 5 x {0.0 kJ mol^-1}]

= -2043.9 kJ mol^-1

We know that,

1 mol = 6.022 x 10²³

  ΔH°_reaction = -2043.9 kJ mol^-1 = -339.41 x 10^-23 kJ = -3.39 x 10^-21 kJ

Hence, ΔH° for this reaction is -3.39 x 10^-21 kJ.

(b) Given,

Number of mole of CO₂ = 3 mol

Number of mole of H₂O = 4 mol

C_CO2(g) = 37.1 J K^-1 mol^-1

C_H2O(g) = 33.6 J K^-1 mol^-1

Since, heat capacity is extensive property. So, it is additive in nature.

Total heat capacity, C = n_CO2(g) x C_CO2(g) + n_H2O(g) x C_H2O(g)

= 3 mol x 37.1 J K^-1 mol^-1 + 4 mol x 33.6 J K^-1 mol^-1

= 245.7 J/K

Hence, the total heat capacity of 3 mol of CO2(g) and 4 mol of H2O(g) is 245.7 J/K.