Question

If enthalpies of formation for $C_2{H}_4\left(g\right)$, $C{O}_2\left(g\right)$ and $H_{2}O\left(l\right)$ at ${25}^o$C and $1$ atm pressure be $52$, $-394$ and $-286$ $kJmo{l}^{-1}$ respectively, then the enthalpy of combustion of $C_2{H}_4\left(g\right)$ will be:

• A. $-141.2$kJ $mo{l}^{-1}$
• B. $-1412$kJ $mo{l}^{-1}$
• C. $+141.2$kJ $mo{l}^{-1}$
• D. $+1412$kJ $mo{l}^{-1}$

Answer 1

The correct option is B $-1412$kJ $mo{l}^{-1}$

Given, $2C\left(g\right)+2H_2\left(g\right)\to C_2{H}_4\left(g\right);\Delta H_1=52kJmo{l}^{-1}$ .......(i)

$C\left(g\right)+O_2\left(g\right)\to C{O}_2\left(g\right);\Delta H_2=-394kJmo{l}^{-1}$ .......(ii)

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right);\Delta H_3-286kJmo{l}^{-1}$ ..........(iii)

Reaction involved for combustion of $C_2{H}_4\left(g\right)$ is,

$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O;\Delta H=?$ .....(iv)

$2\times [Eq.(ii)+Eq.(iii\left)\right]-Eq.\left(i\right),$ we get Eq.(iv)

$\therefore \Delta =2[\Delta H_2+\Delta H_3]-\Delta H_1$

$=2[-394-286]-52$

$=-1360-52=-1412$ $kJmo{l}^{-1}$.

Hence,option B is correct.