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Standard XII

Chemistry

Properties of Enthalpy

The heat of f...

Question

The heat of formation of ethylene is 12.5 kcal. Calculate C=C bond energy in ethylene from the following data. Heat of atomisation of C=170.1 kcal/mole, heat of atomisation of H=52.1 kcal/mole, bond energy of C-H=99.3 kcal/mole.

1.50 k c a l / m o l

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140.5 k c a l / m o l

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130 k c a l / m o l

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160.5 k c a l / m o l

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Solution

The correct option is B $140.5 k c a l / m o l$
$C (s) 2 m o l e ⟶ C (g); Δ H = 2 \times 170.9 k c a l$

$2 H_{2} (g) ⟶ 4 H (g); Δ H = 4 \times 52.1 k c a l$

Given, $2 C (s) + 2 H_{2} (g) ⟶ C_{2} H_{4} (g)$

$Δ H = 2 \times 170.9 + 4 \times 52.1 - 4 \times 99.3 - x$

$12.5 = 2 \times 170.9 + 4 \times 52.1 - 4 \times 99.3 - x$

$x = 140.5 k c a l / m o l e$

Thus, the bond energy of C= C bond = 140.5 kcal/mole

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The heat of formation of ethylene is 12.5 kcal. Calculate C=C bond energy in ethylene from the following data. Heat of atomisation of C=170.1 kcal/mole, heat of atomisation of H=52.1 kcal/mole, bond energy of C-H=99.3 kcal/mole.

The correct option is B 140.5 kcal/molC(s)2 mole⟶C(g); ΔH=2×170.9 kcal 2H2(g)⟶4H(g); ΔH=4×52.1 kcal Given, 2C(s)+2H2(g)⟶C2H4(g) ΔH=2×170.9+4×52.1−4×99.3−x 12.5=2×170.9+4×52.1−4×99.3−x x=140.5 kcal/mole Thus, the bond energy of C= C bond = 140.5 kcal/mole