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Heat of Formation

The enthalpy ...

Question

The enthalpy of formation of $C_{2} H_{2 (g)}$ and $C_{6} H_{6 (g)}$ at $298 K$ and $230$ and $85 k J / m o l$ respectively. The enthalpy change, $Δ H$ for the reaction: $3 C_{2} H_{2 (g)} ⇌ C_{6} H_{6 (g)}$ at $298 K$ is:

A

- 205 k J / m o l

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B

+ 305 k J / m o l

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C

- 605 k J / m o l

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D

+ 605 k J / m o l

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Solution

The correct option is B $+ 605 k J / m o l$
Heat of formation is valid for only one mole of compound,
So For $3$ mole of $C_{2} H_{2}$ , $H_{R} = 3 \times H_{f}$
So $Δ H = H_{R} - H_{p} = 3 \times 230 - 85 = 605$

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