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

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: (A) +890.3 kJ mol -1 (B) -604.5 kJ mol -1 (C) +604.5 kJ mol -1 (D) -890.3 kJ mol -1

Tardigrade Admin · Accepted Answer

2 O 2+ CH 4( g ) arrow CO 2( g )+2 H 2 O ( l ) Δ H =Δ° H f co 2+2 × Δ° H f H2 O -Δ° H f CH 4 =-393.5+2 ×-285.8-(-74.8) =-889.7 kJ / mol ≈-890.3 kJ mol -1

Q. The enthalpy of formation of $C H_{4 (g)}, H_{2} O_{(cl)}$ and $C O_{2 (g)}$ are respectively $- 74.8 k J m o l^{- 1}, - 285.5 k J m o l^{- 1}$ and $- 393.5 k J m o l^{- 1} .$ Then, the standard enthalpy of combustion of $C H_{4 (g)}$ is:

$+ 890.3 k J m o l^{- 1}$

$- 604.5 k J m o l^{- 1}$

$+ 604.5 k J m o l^{- 1}$

$- 890.3 k J m o l^{- 1}$

$2 O_{2} + C H_{4 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)}$

$Δ H = Δ^{\circ} H_{f_{c o_{2}}} + 2 \times Δ^{\circ} H_{f_{H_{2} O}} - Δ^{\circ} H_{f_{C H_{4}}}$

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

$= - 889.7 k J / m o l \approx - 890.3 k J m o l^{- 1}$

Q. The enthalpy of formation of CH4(g)​,H2​O(cl) ​ and CO2(g)​ are respectively −74.8kJmol−1,−285.5kJmol−1 and −393.5kJmol−1. Then, the standard enthalpy of combustion of CH4(g)​ is:

+890.3kJmol−1

−604.5kJmol−1

+604.5kJmol−1

−890.3kJmol−1