Identify the reaction in which the heat liberated corresponds to the heat of formation ( $∆ H$ ):

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$2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O$ (g) $+$ $h e a t$

$C_{(diamond)} + 2 H_{2 (g)} \to {CH}_{4 (g)} +$ $h e a t$

$S_{(r h o m b i c)} + O_{2 (g)} \to S_{2 (g)} + h e a t$

$C_{(diamond)} + O_{2 (g)} \to {CO}_{2 (g)} +$ $h e a t$

answer is D.

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Detailed Solution

Heat of formation states that the enthalpy change during the formation of 1 mole of compound from their constituent elements in the standard state.

We know, the stable state of carbon is graphite. And sulphur mainly exists as monoclinic and rhombic. But the stable state of carbon is rhombic.

Hence, here $S_{(r h o m b i c)} + O_{2 (g)} \to S O_{2 (g)} + h e a t$ follows all the rules of heat of formation and also heat is liberated in this reaction.

Also the product formed is 1 mole and the heat of reaction is equal to the heat of formation.

Therefore, option (D) is correct.

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