Based on the following thermochemical equations: (g) + C(s) - (g) + (g); δ H = 131 kJ; (g) + 1/2 (g) - (g); δ H = -282 kJ; (g) + 1/2 (g) - (g); δ H = -242 kJ; C(s) + (g) - (g); δ H = X kJ. The val

Question

Based on the following thermochemical equations: $H_2O$(g) + C(s) -> $CO$(g) + $H_2$(g); δ H = 131 kJ; $CO$(g) + 1/2 $O_2$(g) -> $CO_2$(g); δ H = -282 kJ; $H_2$(g) + 1/2 $O_2$(g) -> $H_2O$(g); δ H = -242 kJ; C(s) + $O_2$(g) -> $CO_2$(g); δ H = X kJ. The value of 'X' will be

Prashant Kotian · Accepted Answer

Correct answer: A. Adding the first three equations: (g) + C(s) -> (g) + (g) (131) + (g) + 1/2 (g) -> (g) (-282) + (g) + 1/2 (g) -> (g) (-242). This simplifies to C(s) + (g) -> (g). Thus, X = 131 - 282 - 242 = -393 kJ.

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