Question

The standard enthalpies of formation in $ KJ \, mol^{-1} $ for $ CO_{2}(g) $ , $ H_{2}O_{(l)} $ and $ C_{2}H_{5}OH_{(l)} $ are $ -393.5, - 286.0 $ and $ -278.0 $ respectively. The standard enthalpy of combustion of $ C_{2}H_{5}OH_{\left(l\right)} $ in $ kJ \, mol^{-1} $ is

• A. $ -923 $
• B. $ +1367 $
• C. $ +1923 $
• D. $ -1367 $

Answer 1

Answer: (D)

$C_{(s)}+O_{2}(g) \to CO_{2}(g), \Delta\,H =-393.5\, kJ\,mol^{-1} \dots (i)$
$H_{2}(g)+\frac{1}{2}O_{2}(g) \to H_{2}O_{(l)}, \Delta\, H=-393.5\,kJ\,mol^{-1} \dots(ii)$
$2C_{(s)}+3H_{2}(g)+\frac{1}{2}O_{2}(g) \to C_{2}H_{5}OH_{(l)}, \Delta\, H =-278.0\,KJ\,mol^{-1} \dots (iii)$
$C_{2}H_{5}OH_{(l)}+3O_{2}(g) \to 2CO_{2}(g)+3H_{2}O_{(l)}, \Delta\, H=? \dots(iv)$
$(i)\times 2+3\times (ii) - (iii)$ will give equation (iv)
Hence, $\Delta\,H_{combustion}=2(-393.5)+3(-286)-(-278)$
$=-787-858+278$
$=-1367\,KJ\,mol^{-1}$