1. Tardigrade
  2. Question
  3. Chemistry
  4. The enthalpy of formation of CH 4( g ), H 2 O text (cl) and CO 2( g ) are respectively -74.8 kJ mol -1,-285.5 kJ mol -1 and -393.5 kJ mol -1 . Then, the standard enthalpy of combustion of CH 4( g ) is:

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The enthalpy of formation of $CH _{4( g )}, H _{2} O _{\text {(cl) }}$ and $CO _{2( g )}$ are respectively $-74.8\, kJ\, mol ^{-1},-285.5 kJ mol ^{-1}$ and $-393.5\, kJ\, mol ^{-1} .$ Then, the standard enthalpy of combustion of $CH _{4( g )}$ is:

Thermodynamics

Solution:

$2 O _{2}+ CH _{4( g )} \rightarrow CO _{2( g )}+2 H _{2} O _{( l )}$

$\Delta H =\Delta^{\circ} H _{ f _{ co _{2}}}+2 \times \Delta^{\circ} H _{ f _{ H_{2} O }}-\Delta^{\circ} H _{ f _{ CH _{4}}}$

$=-393.5+2 \times-285.8-(-74.8)$

$=-889.7\, kJ / mol \approx-890.3\, kJ\, mol ^{-1}$