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Given ) g ( CO ) g ( O C 2 2 ) graphite ( ⎯→ ⎯ + ; Δ r H° = – 393.5 kJ mol –1 H 2 (g) + 2 1 O 2 (g) ⎯→ H 2 O

1. ; Δ r H° = – 285.8 kJ mol –1 CO 2 (g) + 2H 2 O
2. ⎯→ CH 4 (g) + 2O 2 (g); Δ r H° = + 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) + 2H 2 (g) ⎯→ CH 4 (g) will be ;
3. –74.8 kJ mol –1
4. –144.0 kJ mol –1 (3) +74.8 kJ mol –1 (4) +144.0 kJ mol –1 Students may find similar question in CP exercise sheet : [ JEE Main, Chapter : Chemical Energetic ,Exercise-1 Q. No.12 ] [ JEE Advance, Chapter : Chemical Energetic ,Exercise-4 Q. No.11 ]

Correct answer: ; Δ r H° = – 285.8 kJ mol –1 CO 2 (g) + 2H 2 O

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