For the complete combustion of ethanol, ΔU = q the amount of heat produced as measured in bomb calorimeter is 1364.47 KJ mol-1 at 250C. Assuming ideality, the enthalpy of combustion, ΔHC for the reaction will be [R=8.314JK−1mol−1]

ΔU = -1364.47 KJmol−1 (Amount of heat produced)Enthalpy of combustion is ΔH = ΔU + ΔngRTAccording to equation Δng= 2-3 = -1(C2H5OH(l)+ 302(g)→2 CO2(g)+ 3H2O(l))T=25∘C=298 K thus ;ΔH = -1364 . 47 + [(-1) × 8 . 314 × 2981000]ΔH = -1364 . 47 - 2.477 = - 1366.947 KJ mol-1

For the complete combustion of ethanol, ΔU = q the amount of heat produced as measured in bomb calorimeter is 1364.47 KJ mol-1 at 250C. Assuming ideality, the enthalpy of combustion, ΔHC for the reaction will be [R=8.314JK−1mol−1]

ΔU = -1364.47 KJmol−1 (Amount of heat produced)Enthalpy of combustion is ΔH = ΔU + ΔngRTAccording to equation Δng= 2-3 = -1(C2H5OH(l)+ 302(g)→2 CO2(g)+ 3H2O(l))T=25∘C=298 K thus ;ΔH = -1364 . 47 + [(-1) × 8 . 314 × 2981000]ΔH = -1364 . 47 - 2.477 = - 1366.947 KJ mol-1

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For the complete combustion of ethanol, $ΔU = q$ the amount of heat produced as measured in bomb calorimeter is $1364.47 K J m o l^{- 1}$ at 25⁰C. Assuming ideality, the enthalpy of combustion, ${ΔH}_{C}$ for the reaction will be $[R = 8.314 {JK}^{- 1} {mol}^{- 1}]$

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$- 1366.95 KJ {mol}^{- 1}$

$- 1460.50 KJ {mol}^{- 1}$

$- 1361.95 KJ {mol}^{- 1}$

$- 1350.50 KJ {mol}^{- 1}$

answer is A.

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Detailed Solution

${ΔU = -1364.47 KJmol}^{- 1}$ (Amount of heat produced)
Enthalpy of combustion is
${ΔH = ΔU + Δn}_{g} RT$
According to equation ${Δn}_{g} = 2-3 = -1$
$(C_{2} H_{5} {OH}_{(l)} + 30_{2_{(g)}} \to {2 CO}_{2_{(g)}} {+ 3H}_{2} O_{(l)})$
$T = 25^{\circ} C = 298 K$ thus ;
$ΔH = -1364 . 47 + [\frac{(-1) × 8 . 314 × 298}{1000}]$
${ΔH = -1364 . 47 - 2.477 = - 1366.947 KJ mol}^{-1}$