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Heat of Reaction

Calculate H r...

Question

Calculate $△ H_{r}^{o}$ for the reaction $C H_{2} C l_{2} (g) \to C (g) + 2 H (g) + 2 C l (g) .$
The average bond enthalpies of C - H and C - Cl bonds are $414 k J m o l^{- 1}$ and $330 k J m o l^{- 1} .$

1488 k J m o l^{- 1}

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1468 k J m o l^{- 1}

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1478 k J m o l^{- 1}

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1498 k J m o l^{- 1}

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Solution

The correct option is A $1488 k J m o l^{- 1}$
We are given
$ε_{C - H} = 414 k J m o l^{- 1}$
$ε_{C - C l} = 330 k J m o l^{- 1}$

We know
$Δ_{r} H = \sum (B o n d E n e r g y)_{r e a c t a n t s} - \sum (B o n d E n e r g y)_{p r o d u c t s}$
Here the bond energy of products won't be taken into account as the products doesn't contain any bonds.
So,
$△ H_{r} = 2 ε_{C - H} + 2 ε_{C - C l}$
putting the values,
$= 2 \times 414 + 2 \times 330$
$= 828 + 660 = 1488 k J / m o l .$
hence, $△ H_{r} = 1488 k J / m o l .$

Theory:

Determination of enthalpy of reaction:
$Δ_{r} H = \sum (B o n d E n e r g y)_{r e a c t a n t s} - \sum (B o n d E n e r g y)_{p r o d u c t s}$

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Heat of Reaction

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Calculate △Hor for the reaction CH2Cl2(g)→C(g)+2H(g)+2Cl(g). The average bond enthalpies of C - H and C - Cl bonds are 414 kJmol−1 and 330 kJ mol−1.

Calculate $△ H_{r}^{o}$ for the reaction $C H_{2} C l_{2} (g) \to C (g) + 2 H (g) + 2 C l (g) .$
The average bond enthalpies of C - H and C - Cl bonds are $414 k J m o l^{- 1}$ and $330 k J m o l^{- 1} .$