Question

The amount of heat evolved for the combustion of ethane is -900kJ mol^-1 at 300K and 1 atm.

C₂H_6(g) + \(\frac{7}{2}\)O_2(g) → 2CO_2(g)  + 3H₂O_(I)

Calculate W, ΔH and ΔU for the combustion of 12 × 10^-3 kg ethane.

Answer 1

Given : 

ΔH = -900 kJ mol^-1

Temperature = T = 300 K 

Pressure = P = 1 atm 

Mass of ethane = m = 12 × 10^-3 kg

Molar mass of ethane (C₂H₆) = 30 × 10 kg mol^-1

ΔH = ? 

ΔU = ? for given ethane.

Number of moles of C₂H₆ = n = \(\frac{m\,kg}{M\,kg\,mol^{-1}}\)

= \(\frac{12\times 10^{-3}}{30\times 10^{-3}}\)

= 0.4 mol 

For the given reaction,

C₂H_6(g) + \(\frac{7}{2}\)O_2(g) → 2CO_2(g)  + 3H₂O_(I)

Δn = (n₂)_{gaseous products} - (n₁)_{gaseous reactants}

= 2 – (1 + \(\frac{7}{2}\)) 

= -2.5 mol

For 1 mol of C₂H₆,

 Δn = -2.5 mol

∴ For 0.4 mol of C₂H₆,

Δn = -2.5 × 0.4

= -1 mol

Since there is a decrease in number of moles, the work is of compression on the system.

W = -ΔnRT 

= – (-1) × 8.314 × 300 

= + 2494 J 

= + 2.494 kJ

For 1 mol of C₂H₆,

ΔH = -900 kJ

∴ For 0.4 mol of C₂H₆, 

ΔH = – 900 × 0.4

= – 360 kJ

ΔH = ΔU + ΔnRT

ΔU = ΔH – ΔnRT

= -360 – (-1) × 8.314 × 300× 10^-3

= – 360 + 2.494

= – 357.506 kJ