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Chemistry Question on Thermodynamics

For the complete combustion of ethanol, C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l),C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}, the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ moI−11364.47 \,kJ \,moI^{-1} at 25∘25^{\circ}C. Assuming ideality the enthalpy of combustion, ΔCH\Delta_ CH, for the reaction will be (R=8.314 J K−1(R = 8.314\, J \,K^{-1} mol−1)mol^{-1})

A

−1366.95 kJ moI−1- 1366.95 \,kJ \,moI^{-1}

B

−1361.95 kJ moI−1- 1361.95 \, kJ \, moI^{-1}

C

−1460.50 kJ moI−1- 1460.50 \, kJ \,moI^{-1}

D

−1350.50 kJ moI−1- 1350.50\, kJ \, moI^{-1}

Answer

−1366.95 kJ moI−1- 1366.95 \,kJ \,moI^{-1}

Explanation

Solution

C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)C _{2} H _{5} OH (l)+3 O _{2}( g ) \to 2 CO _{2}( g )+3 H _{2} O (l) Bomb calorimeter gives Δ U\Delta \,U of the reaction So, as per question ΔU=−1364.47 kJ mol−1\Delta U =-1364.47 \,kJ \,mol ^{-1} Δgg=−1\Delta g _{ g } =-1 Δ H=Δ U+ΔngRT\Delta \,H =\Delta\, U +\Delta n _{ g } RT =−1364.47−1×8.314×2981000=-1364.47-\frac{1 \times 8.314 \times 298}{1000} =−1366.93 kJ mol−1=-1366.93\, kJ \,mol ^{-1}