Identify the reaction in which the heat liberated corresponds to the heat formation $(Δ H)$ .

C (d i a m o n d) + O_{2} \to {C O}_{2} (g) + h e a t

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

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C (d i a m o n d) + 2 H_{2} (g) \to {C H}_{4} (g) + h e a t

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

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Solution

The correct option is C $S (r h o m b i c) + O_{2} (g) \to {S O}_{2} (g) + h e a t$
Heat of formation is defined as the energy involved in the formation of 1 mole of that species from its constituents elements.
Since, in option $A$ and $C$ the products form is 1 mole only but the constituent elements taken are ${C(diamond) and H}_{2}$ . But, for carbon the chosen element is $C (G r a p h i t e)$ as constituent element.
So, $A$ and $C$ are not the formation reaction
Hence, their heat of reaction is not equal to heat of formation.
In $B$ , $2 moles$ of $H_{2} O$ is formed not $1$ .
Hence, it is also not formation reaction.
In option $D$ , product formed is $1 mole$ and also the reactants chosen are constituent elements so its a formation reaction for $S O_{2}$ .
Hence, its heat of reaction is equal to heat of formation.
Hence, the answer is option $D$ .

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Similar questions

C_{(s)} + O_{2 (g)} \to C O_{2 (g)};

Δ H = - 394 k J

2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(l)};

Δ H = - 568 k J

C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)};

Δ H = - 394 k J

C H_{4}

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (g)

Δ H = - 573.2 k J

Assertion- $C (g r a p h i t e) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 393.5 k . J$

$C (d i a m o n d) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 395.4 k . J$

Reason(R): graphite is more stable than diamond, so heat of combustion is less.

C_{D i a m o n d} + O_{2} \to C O_{2} (g); Δ H = - 94.3 k c a l

C_{G r a p h i t e} + O_{2} \to C O_{2} (g); Δ H = - 97.6 k c a l

C_{d i a m o n d} \to C_{g r a p h i t e}

C_{(d i a m o n d)} + O_{2} (g) \to C O_{2} (g) △ H = - 94.5 K c a l

C_{(g r a p h i t e)} + O_{2} (g) \to C O_{2} (g) △ H = - 94.0 K c a l

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