$0.46 g$ of liquid ethanol ( $M = 46 g mo l^{- 1}$ ) is burnt completely in a bomb calorimeter whose heat capacity (including contents) is $10.4 kJ K^{- 1}$ ; the temperature rises by $1.31 K$ . For $C X_{2} H X_{5} OH (l) + 3 O X_{2} (g) 2 CO X_{2} (g) + 3 H X_{2} O (l)$ at $298 K$ find the molar enthalpy of combustion in $kJ mo l^{- 1}$ (nearest integer). Use $R = 8.314 J mo l^{- 1} K^{- 1}$ .

Numerical answer type, answer in kJ/mol, tolerance 1.

Answer: -1365 kJ/mol

Heat released

= 10.4 \times 1.31 = 13.62 kJ

for

0.010 mol

, so

Δ_{c} U = - 1362.4 kJ mo l^{- 1}

. For this reaction (liquid water)

Δ n_{g} = 2 - 3 = - 1

, so

Δ_{c} H = Δ_{c} U + Δ n_{g} RT = - 1362.4 - 2.48 = - 1364.9 kJ mo l^{- 1}

, about -1365.