Question

What will be the enthalpy of combustion of ethylene (gas) to form $C{O}_2$ (gas) and water at $298K$ and $1atm$ pressure? The enthalpies of formation of $C{O}_2,H_{2}O$ and $C_2{H}_4$ are $-393.5,-241.8,+52.3kJ/mol$ respectively.

• A. $-1332.9kJmo{l}^{-1}$
• B. $-1322.9kJmo{l}^{-1}$
• C. $+1332.9kJmo{l}^{-1}$
• D. $+1322.9kJmo{l}^{-1}$

Answer 1

Answer: (B)

$C_2{H}_4(g)+3O_2(g)\to 2C{O}_2(g)+2H_{2}O(g)$
Given, ${\Delta }_f{H}_{\left(C{O}_2\right)}^o=-393.5kJmo{l}^{-1}$
${\Delta }_f{H}_{\left(H_{2}O\right)}^{\circ }=-241.8kJmo{l}^{-1}$
${\Delta }_f{H}_{\left(C_2{H}_4\right)}^{\circ }=+52.3kJmo{l}^{-1}$
${\Delta }_r{H}^{\circ }$
$=$ [sum of ${\Delta }_f{H}^o$ values of products] -[sum of ${\Delta }_f{H}^o$ values of reactants]
$=\left[2\times {\Delta }_f{H}_{\left(C{O}_2\right)}^{\circ }+2\times {\Delta }_f{H}_{\left(H_{2}O\right)}^{\circ }\right]$
$-\left[{\Delta }_f{H}_{\left(C_2{H}_4\right)}^{\circ }+3\times {\Delta }_f{H}_{\left(O_2\right)}^{\circ }\right]$
$=[2\times (-393.5)+2\times (-241.8)]-[(52.3)+0]$
$(\because {\Delta }_f{H}^{\circ }$ for elementary substance $=0)$
$=[-787.0-486.6]-52.3$
$=-1322.9kJmo{l}^{-1}$