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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

270 k J m o l^{- 1}

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70 k J m o l^{- 1}

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200 k J m o l^{- 1}

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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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The enthalpies of atomization of CH4(g) and C2H6(g) are respectively 400 kJ mol−1 and 670 kJ mol−1. The value of △HC−C would be

The correct option is B 70 kJ mol−1CH4(g)→C(g)+4H(g) △Hatom=400 kJ △Hatom=4×ΔHC−H ∴ △HC−H=4004=100 kJ mol−1 CH3−CH3(g)→2C(g)+6H(g); △Hatom=670 kJ 670=△HC−C+6△HC−H ∴△HC−C=670−6×100=70 kJ mol−1 hence, △HC−C=70 kJ mol−1

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