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Heat of Reaction

The enthalpie...

Question

The enthalpies of atomization of $C H_{4} (g)$ and $C_{2} H_{6} (g)$ are respectively $400 k J m o l^{- 1}$ and $670 k J m o l^{- 1} .$ The value of $△ H_{C - C}$ would be

A

270 k J m o l^{- 1}

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B

70 k J m o l^{- 1}

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C

200 k J m o l^{- 1}

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D

240 k J m o l^{- 1}

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Solution

The correct option is B $70 k J m o l^{- 1}$
$C H_{4} (g) \to C (g) + 4 H (g) △ H_{a t o m} = 400 k J$
$△ H_{a t o m} = 4 \times Δ H_{C - H}$
$∴ △ H_{C - H} = \frac{400}{4} = 100 k J m o l^{- 1}$
$C H_{3} - C H_{3} (g) \to 2 C (g) + 6 H (g); △ H_{a t o m} = 670 k J$
$670 = △ H_{C - C} + 6 △ H_{C - H}$
$∴ △ H_{C - C} = 670 - 6 \times 100 = 70 k J m o l^{- 1}$
hence, $△ H_{C - C} = 70 k J m o l^{- 1}$

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Heat of Reaction

Standard XII Chemistry

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