For the complete combustion of ethanol, C2H5OH(l) + 3O2(g) arrow 2CO2(g) + 3H2O(l), the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol-1 at 25°C. Assuming ideality the enthalpy of combustion, Δ CH, for the reaction will be (R = 8.314 J K-1 mol-1)

Question

For the complete combustion of ethanol, C2H5OH(l) + 3O2(g) arrow 2CO2(g) + 3H2O(l), the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol-1 at 25°C. Assuming ideality the enthalpy of combustion, Δ CH, for the reaction will be (R = 8.314 J K-1 mol-1) (A) - 1366.95 kJ mol-1 (B) - 1361.95 kJ mol-1 (C) - 1460.50 kJ mol-1 (D) - 1350.50 kJ mol-1

Tardigrade Admin · Accepted Answer

C 2 H 5 OH (l)+3 O 2(g) longrightarrow 2 CO 2(g)+3 H 2 O (l) Δ U=-1364.47 kJ / mol Δ H=Δ U+Δ ng R T Δ ng=-1 Δ H=-1364.47+(-1 × 8.314 × 298/1000) [Here, value of R in unit of J must be converted into kJ =-1364.47-2.4776=-1366.9476 kJ / mol or =-1366.95 kJ / mol

$- 1366.95 k J m o l^{- 1}$

$- 1361.95 k J m o l^{- 1}$

$- 1460.50 k J m o l^{- 1}$

$- 1350.50 k J m o l^{- 1}$

Q. For the complete combustion of ethanol, C2​H5​OH(l)​+3O2(g)​→2CO2(g)​+3H2​O(l)​, the amount of heat produced as measured in bomb calorimeter, is 1364.47kJmol−1 at 25∘C. Assuming ideality the enthalpy of combustion, ΔC​H, for the reaction will be (R=8.314JK−1 mol−1)

−1366.95kJmol−1

−1361.95kJmol−1

−1460.50kJmol−1

−1350.50kJmol−1

C2​H5​OH(l)+3O2​(g)⟶2CO2​(g)+3H2​O(l) ΔU=−1364.47kJ/mol ΔH=ΔU+Δng​RT Δng​=−1 ΔH=−1364.47+1000−1×8.314×298​ [Here, value of R in unit of J must be converted into kJ =−1364.47−2.4776=−1366.9476kJ/mol or =−1366.95kJ/mol

$- 1366.95 k J m o l^{- 1}$

$- 1361.95 k J m o l^{- 1}$

$- 1460.50 k J m o l^{- 1}$

$- 1350.50 k J m o l^{- 1}$