Given C text(graphite) + O2 (g) → CO2 (g) ; Δr H° = - 393.5 kJ mol-1 H2(g) + (1/2) O2 (g) → H2 O(l); Δr H° = - 285.8 kJ mol-1 CO2(g) + 2H2 O(l) - CH4(g) + 2O2(g) ; Δr H° = + 890.3 kJ mol-1 Based on the above thermochemical equations, the value of Δr H° at 298 K for the reaction C text(graphite) + 2H2 (g) → CH4(g) will be :

Question

Given C text(graphite) + O2 (g) → CO2 (g) ; Δr H° = - 393.5 kJ mol-1 H2(g) + (1/2) O2 (g) → H2 O(l); Δr H° = - 285.8 kJ mol-1 CO2(g) + 2H2 O(l) -> CH4(g) + 2O2(g) ; Δr H° = + 890.3 kJ mol-1 Based on the above thermochemical equations, the value of Δr H° at 298 K for the reaction C text(graphite) + 2H2 (g) → CH4(g) will be : (A) - 74.8 kJ mol-1 (B) -144.0 kJ mol-1 (C) + 74.8 kJ mol-1 (D) + 144.0 kJ mol-1

Tardigrade Admin · Accepted Answer

C text (graphite) + O 2( g ) arrow CO 2( g ) Δ r H° =-393.5 kJ mol -1...(i) H 2( g )+(1/2) O 2( g ) arrow H 2 O ( I ) Δr H°=-285.8 kJ mol -1...(ii) CO 2( g )+2 H 2 O ( I ) arrow CH 4( g )+2 O 2( g) Δ r H ° =890.3 kJ mol -1 ...(iii) By applying the operation (i) +2 × (ii )+ (iii), we get C text (graphite) +2 H 2( g ) arrow CH 4( g ) Δr H° =-393.5-285.8 × 2+890.3 =-74.8 kJ mol -1

$- 74.8 k J m o l^{- 1}$

$- 144.0 k J m o l^{- 1}$

$+ 74.8 k J m o l^{- 1}$

$+ 144.0 k J m o l^{- 1}$

−74.8kJmol−1

−144.0kJmol−1

+74.8kJmol−1

+144.0kJmol−1

$- 74.8 k J m o l^{- 1}$

$- 144.0 k J m o l^{- 1}$

$+ 74.8 k J m o l^{- 1}$

$+ 144.0 k J m o l^{- 1}$