Question

Calculate the work (w) and ΔEo, in kJ, at 298 K and 1 atm pressure, for the combustion of one mole of C4H10 (g). First write and balance the equation. The products will be CO2 (g) and H2O (g)
The value of ΔHo for this reaction is -2658.3 kJ/mol

Answer 1

Answer:

Explanation:

2C₄H₁₀ + 13O₂ = 8 CO₂ + 10H₂O

Change in number of moles Δn = 18 - 15 = + 3 moles .

ΔHo = -2658.3 kJ/mol.

ΔHo = ΔEo+ Δn RT

Δn = 3

For one mole Δn = 1.5

ΔHo = ΔEo+ W

W = Δn RT

= 1.5 x 8.31 x 298

= 3714.5 J

= 3.7 kJ /mole

ΔHo = ΔEo+ W

ΔEo =  ΔHo -  W

= -2658.3 - 3.7  kJ

= - 2662 kJ .

Answer 2

(a) The work done for the combustion of 1 mole of methane is 3,714.5 J.

(b) The change in the energy of the given system is -3,991.2 kJ.

The balanced chemical reaction of the given compounds is written as follows;

$2C_4H_{10} \ + \ 13O_2 \ -->\ 8CO_2 \ + \ 10H_2 O$

The change in number of moles;

$\Delta n = 18 - 15 = 3$

The work done for the combustion of 1 mole of methane is calculated as follows;

$W = \frac{\Delta n}{2} RT\\\\W = \frac{3}{2} \times 8.31 \times 298\\\\W = 3,714.6 \ J$

The energy change is calculated as follows;

$\Delta E_0 = \Delta H_o - W\\\\\Delta E_0 = (\frac{3 \ mole}{2} \times-2658.3 \ kJ/mol) - 3.715 \ kJ\\\\\Delta E_0 = -3,991.2 \ kJ$

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