Question

If enthalpies of formation for $C_{2}H_{4}\left(g\right), CO_{2}\left(g\right)g$ and $H_{2}O\left(l\right)$ at $25^{\circ}\, C$ and 1 atm pressure be $52, -394$ and $-286\, kJ\, mol^{-1}$ respectively, enthalpy of combustion of $C_2H_4\, (g)$ will be

• A. $+141.2\, kJ\, mol^{-1}$
• B. $+1412\, kJ\, mol^{-1}$
• C. $-141.2\, kJ\, mol^{-1}$
• D. $-1412\, kJ\, mol^{-1}$

Answer 1

Answer: (D)

Given,

$2 C (g)+2 H _{2}(g) \longrightarrow C _{2} H _{4}(g)$

$\Delta H_{1}=52 kJ mol ^{-1}$ ...(i)

$C (g)+ O _{2}(g) \longrightarrow CO _{2}(g)$

$\Delta H_{2}=-394 kJ mol ^{-1}$ ...(ii)

$H _{2}(g)+\frac{1}{2} O _{2}(g) \longrightarrow H _{2} O (g)$

$\Delta H_{3}=-286 kJ mol ^{-1}$ ...(iii)

Reaction involved for combustion of $C _{2} H _{4}(g)$ is,

$C _{2} H _{4}+3 O _{2} \longrightarrow 2 CO _{2}+2 H _{2} O ; \Delta H=?$ ...(iv)

$2 \times[\text { Eq. (ii) + Eq. (iii)] - Eq. (i), we get Eq. (iv) }$

$\therefore \Delta H =2\left[\Delta H_{2}+\Delta H_{3}\right]-\Delta H_{1}$

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

$=-1360-52=-1412\, kJ\, mol ^{-1}$