If enthalpies of formation for C2H4(g), CO2(g)g and H2O(l) at 25° C and 1 atm pressure be 52, -394 and -286 kJ mol-1 respectively, enthalpy of combustion of C2H4 (g) will be

Question

If enthalpies of formation for C2H4(g), CO2(g)g and H2O(l) at 25° C and 1 atm pressure be 52, -394 and -286 kJ mol-1 respectively, enthalpy of combustion of C2H4 (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

Tardigrade Admin · Accepted Answer

Given, 2 C (g)+2 H 2(g) longrightarrow C 2 H 4(g) Δ H1=52 kJ mol -1 ...(i) C (g)+ O 2(g) longrightarrow CO 2(g) Δ H2=-394 kJ mol -1 ...(ii) H 2(g)+(1/2) O 2(g) longrightarrow H 2 O (g) Δ H3=-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 ; Δ H=? ...(iv) 2 ×[ text Eq. (ii) + Eq. (iii)] - Eq. (i), we get Eq. (iv) ∴ Δ H =2[Δ H2+Δ H3]-Δ H1 =2[-394-286]-52 =-1360-52=-1412 kJ mol -1

Q. If enthalpies of formation for $C_{2} H_{4} (g), C O_{2} (g) g$ and $H_{2} O (l)$ at $2 5^{\circ} C$ and 1 atm pressure be $52, - 394$ and $- 286 k J m o l^{- 1}$ respectively, enthalpy of combustion of $C_{2} H_{4} (g)$ will be

$+ 141.2 k J m o l^{- 1}$

$+ 1412 k J m o l^{- 1}$

$- 141.2 k J m o l^{- 1}$

$- 1412 k J m o l^{- 1}$

Q. If enthalpies of formation for C2​H4​(g),CO2​(g)g and H2​O(l) at 25∘C and 1 atm pressure be 52,−394 and −286kJmol−1 respectively, enthalpy of combustion of C2​H4​(g) will be

+141.2kJmol−1

+1412kJmol−1

−141.2kJmol−1

−1412kJmol−1

Q. If enthalpies of formation for $C_{2} H_{4} (g), C O_{2} (g) g$ and $H_{2} O (l)$ at $2 5^{\circ} C$ and 1 atm pressure be $52, - 394$ and $- 286 k J m o l^{- 1}$ respectively, enthalpy of combustion of $C_{2} H_{4} (g)$ will be

$+ 141.2 k J m o l^{- 1}$

$+ 1412 k J m o l^{- 1}$

$- 141.2 k J m o l^{- 1}$

$- 1412 k J m o l^{- 1}$