Question

What will be the enthalpy of combustion of ethylene (gas) to form $CO _{2}$ (gas) and water at $298 \,K$ and $1\, atm$ pressure? The enthalpies of formation of $CO _{2}, H _{2} O$ and $C_{2} H_{4}$ are $-393.5,-241.8,+52.3\, kJ /mol$ respectively.

• A. $-1332.9 \,kJ \,mol ^{-1}$
• B. $ -1322.9\,kJ\,mol^{-1} $
• C. $ +1332.9 \,kJ\,mol^{-1} $
• D. $ +1322.9\,kJ\,mol^{-1}$

Answer 1

Answer: (B)

$C _{2} H _{4}(g)+3 O _{2}(g) \rightarrow 2 CO _{2}(g)+2 H _{2} O (g)$
Given, $\Delta_{f} H_{\left( CO _{2}\right)}^{o}=-393.5\, kJ \,mol ^{-1}$
$\Delta_{f} H_{\left(H_{2} O \right)}^{\circ}=-241.8 \,kJ \,mol ^{-1}$
$\Delta_{f} H_{\left(C_{2} H_{4}\right)}^{\circ}=+52.3\, kJ \,mol ^{-1}$
$\Delta_{r} H^{\circ}$
$=$ [sum of $\Delta_{f} H^{o}$ values of products] -[sum of $\Delta_{f} H^{o}$ values of reactants]
$=\left[2 \times \Delta_{f} H_{\left(C O_{2}\right)}^{\circ}+2 \times \Delta_{f} H_{\left(H_{2} O\right)}^{\circ}\right]$
$-\left[\Delta_{f} H_{\left(C_{2} H_{4}\right)}^{\circ}+3 \times \Delta_{f} H_{\left(O_{2}\right)}^{\circ}\right]$
$=[2 \times(-393.5)+2 \times(-241.8)]-[(52.3)+0]$
$\left(\because \Delta_{f} H^{\circ}\right.$ for elementary substance $\left.=0\right)$
$=[-787.0-486.6]-52.3$
$=-1322.9\, k J \,mol ^{-1}$