Given: C (graphite )+ O 2( g ) arrow CO 2( g ) ; Δ r H o =-393.5 kJ mol -1 H 2( g )+(1/2) O 2( g ) arrow H 2 O ( l ) ; Δ r H 0=-285.8 kJ mol -1 CO 2( g )+2 H 2 O ( l ) arrow CH 4( g )+2 O 2( g ) ; Δ r H 0=+890.3 kJ mol -1 Based on the above thermochemical equations, the value of Δ r H ° at 298 K for the reaction C (graphite )+2 H 2( g ) arrow CH 4( g ) will be:

Question

Given: C (graphite )+ O 2( g ) arrow CO 2( g ) ; Δ r H o =-393.5 kJ mol -1 H 2( g )+(1/2) O 2( g ) arrow H 2 O ( l ) ; Δ r H 0=-285.8 kJ mol -1 CO 2( g )+2 H 2 O ( l ) arrow CH 4( g )+2 O 2( g ) ; Δ r H 0=+890.3 kJ mol -1 Based on the above thermochemical equations, the value of Δ r H ° at 298 K for the reaction C (graphite )+2 H 2( g ) arrow CH 4( g ) will be: (A) +144.0 kJ mol- 1 (B) -74.8 kJ mol- 1 (C) -144.0 kJ mol- 1 (D) +74.8 kJ mol- 1

Tardigrade Admin · Accepted Answer

<blockquote> The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions. The enthalpy of combustion of a substance is defined as the heat energy given out when one mole of a substance burns completely in oxygen. The formation reaction of methane is C(graphite)+2H2(g) arrow (CH)4(g) The heat of combustion data for C (graphite), H 2( g ) and CH 4( g ) beginarrayl Δ H f o =∑ Δ H text Combustion ( text reactants )-∑ Δ H text Combustion text (products) Δ H f o ( CH 4)=Δ H C o ( C text (graphite )+2 × Δ H C o ( H 2( g ))-Δ H C o ( CH 4( g )) Δ H f o ( CH 4)=-393.5-2 × 285.8+890.3 Δ H o text of CH 4 ( g )=-74.8 kJ / mol endarray

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

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

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

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

+144.0kJmol−1

−74.8kJmol−1

−144.0kJmol−1

+74.8kJmol−1

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

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

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

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