For complete combustion of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\ell)$, the amount of heat produced as measured in bomb calorimeter, is $1364.47 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $25^{0} \mathrm{C}$. Assuming ideality the Enthalpy of combustion, $\Delta_{\mathrm{C}} \mathrm{H}$, for the reaction will be: $\left(\mathrm{R}=8.314 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

(C) $-1366.95 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(A) $-1460.50 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(B) $-1350.50 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(D) $-1361.95 \mathrm{~kJ} \mathrm{~mol}^{-1}$

JEE Main · 2014 · 06 Apr 2014 · Offline · Chemical Thermodynamics

For complete combustion of ethanol,

C_{2} H_{5} OH (ℓ) + 3 O_{2} (g) ⟶ 2 CO_{2} (g) + 3 H_{2} O (ℓ)

, the amount of heat produced as measured in bomb calorimeter, is

1364.47 kJ mol^{- 1}

2 5^{0} C

. Assuming ideality the Enthalpy of combustion,

Δ_{C} H

, for the reaction will be:

(R = 8.314 kJ mol^{- 1})

Answer: (C)

A bomb calorimeter measures the heat at constant volume, which is the internal-energy change

Δ E

; the enthalpy of combustion needs the constant-pressure value through

Δ H = Δ E + Δ n_{g} R T

. Only gases count in

Δ n_{g}

, and here three moles of oxygen give two moles of carbon dioxide while the water is liquid, so

Δ n_{g} = 2 - 3 = - 1

. With

Δ E = - 1364.47

kJ/mol,

T = 298

K and

R = 8.314 \times 1 0^{- 3}

kJ mol

^{- 1}

^{- 1}

Δ H = - 1364.47 + (- 1) (8.314 \times 1 0^{- 3}) (298) = - 1364.47 - 2.48 = - 1366.95

kJ/mol.

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option