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Sonbull [250]
3 years ago
7

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
Chemistry
1 answer:
Paul [167]3 years ago
3 0

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 .

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The ideal gas law may be written as
p= \frac{\rho R T}{M}
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)

For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
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Therefore
p= \frac{(0.09 \, kg/m^{3})*(8.314 \, J/(mol-K))*(299 \, K)}{1.008 \times 10^{-3} \, kg/mol} =2.2195 \times 10^{5} \, Pa

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Therefore
p = 2.2195 x 10⁵ Pa
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Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)

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Please provide the choices here.

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