Question

the enthalpy of formation of C2H5OH ,CO2 and H2O are -277 ,393 and -285.5 kj/mol calculate the enthaloy change for the reaction C2H5OH + 3O2 = 2CO2+3H2O

Answer 1

Answer:

$\large \boxed{\text{-1366 kJ/mol}}$

Explanation:

C₂H₅OH + 3O₂ ⟶ 2CO₂ + 3H₂O

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

                            C₂H₅OH + 3O₂ ⟶ 2CO₂ + 3H₂O

ΔH°f/kJ·mol⁻¹:         -277                     -393    -285.5

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & 2(-393) + 3(-285.5) -(-277)\\& = & -786 - 856.5 + 277\\& = & \textbf{-1366 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-1366 kJ/mol}}$