If enthalpies of formation for C2H4(g), CO2(g) and H2O (l) at 25oC and 1 atm pressure are 52, -394 and -286 kJ/mol respectively, then total enthalpy of combustion is

Question

If enthalpies of formation for C2H4(g), CO2(g) and H2O (l) at 25oC and 1 atm pressure are 52, -394 and -286 kJ/mol respectively, then total enthalpy of combustion is (A)  +1412 kJ/mol  (B)  +141.2 kJ/mol  (C)  -141.2 kJ/mol  (D)  -14.2 kJ/mol

Tardigrade Admin · Accepted Answer

Given: Enthalpies of C2H4(g), CO2(g) and H2O (l) 52, -394 and -286 respectively. We know that C2H4+3O2 xrightarrow2CO2+2H2O Therefore, total enthalpy of their combustion =2× (-394)+2× (-286)-(52) =-1412 kJ/mol

Q. If enthalpies of formation for $C_{2} H_{4} (g), C O_{2} (g)$ and $H_{2} O (l)$ at $25^{o} C$ and $1 a t m$ pressure are $52,$ $- 394$ and $- 286 k J / m o l$ respectively, then total enthalpy of combustion is

$+ 1412 k J / m o l$

$+ 141.2 k J / m o l$

$- 141.2 k J / m o l$

$- 14.2 k J / m o l$

Given: Enthalpies of $C_{2} H_{4} (g), C O_{2} (g)$ and $H_{2} O (l)$ $52, - 394$ and $- 286$ respectively. We know that $C_{2} H_{4} + 3 O_{2} 2 C O_{2} + 2 H_{2} O$ Therefore, total enthalpy of their combustion $= 2 \times (- 394) + 2 \times (- 286) - (52)$ $= - 1412 k J / m o l$

Q. If enthalpies of formation for C2​H4​(g),CO2​(g) and H2​O(l) at 25oC and 1atm pressure are 52, −394 and −286kJ/mol respectively, then total enthalpy of combustion is

+1412kJ/mol

+141.2kJ/mol

−141.2kJ/mol

−14.2kJ/mol