1. Tardigrade
  2. Question
  3. Chemistry
  4. 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 :

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Q. Given
$C_{\text{(graphite)} }+ O_{2 (g)} \to CO_{2 (g)}$ ;
$\Delta_r H^{\circ} = - 393.5 \, kJ \, mol^{-1}$
$H_2(g) + \frac{1}{2} O_2 (g) \to H_2 O(l)$;
$\Delta_r H^{\circ} = - 285.8 \, kJ \, mol^{-1}$
$CO_2(g) + 2H_2 O(l) -> CH_4(g) + 2O_2(g) $;
$\Delta_r H^{\circ} = + 890.3 \, kJ \, mol^{-1}$
Based on the above thermochemical equations, the value of $\Delta_{r} H^{\circ}$ at $298 \,K$ for the reaction
$C_{\text{(graphite)}} + 2H_{2 (g)} \to CH_{4(g) } $ will be :

JEE MainJEE Main 2017Thermodynamics

Solution:

$C _{\text {(graphite) }}+ O _{2}( g ) \rightarrow CO _{2}( g )$

$\Delta_{ r } H^{\circ} =-393.5\, kJ\, mol ^{-1}$...(i)

$H _{2}( g )+\frac{1}{2} O _{2}( g ) \rightarrow H _{2} O ( I )$

$\Delta_{r} H^{\circ}=-285.8\, kJ\, mol ^{-1}$...(ii)

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

$\Delta_{ r } H ^{\circ} =890.3\, kJ\, mol ^{-1}$ ...(iii)

By applying the operation

$ (i) +2 \times (ii )+ (iii),$ we get

$C_{\text {(graphite) }}+2 H _{2}( g ) \rightarrow CH _{4}( g )$

$\Delta_{r} H^{\circ} =-393.5-285.8 \times 2+890.3$

$=-74.8\, kJ\, mol ^{-1}$