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Heat of Reaction

The reaction ...

Question

The reaction of 1 mole of benzene take place at $298 K$ and $1 a t m$ as-
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$
in this reaction $C O_{2} (g)$ and $H_{2} O (l)$ are produced and 3267.0 $k J$ heat is liberated. calculate the $Δ_{f} H^{o}$ for benzene.
$Δ_{f} H^{o}$ for $C O_{2} (g)$ and $H_{2} O (l)$ are $- 393.5 k J m o l^{- 1}$ and $- 285.83 k J m o l^{- 1}$ respectively.

A

68.51 J m o l^{- 1}

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B

- 70.51 J m o l^{- 1}

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C

55.51 k J m o l^{- 1}

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D

48.51 k J m o l^{- 1}

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Solution

The correct option is D $48.51 k J m o l^{- 1}$
Given, $Δ_{f} H^{o} C O_{2} (g) = - 393.5 k J m o l^{- 1}, Δ_{f} H^{o} H_{2} O (l) = - 285.83 k J m o l^{- 1}$

Reaction is:
$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)$
$Δ_{r} H = - 3267.0 k J$

so,
$Δ_{f} H^{o}$ for benzene $= 6 Δ_{f} H^{o} (C O_{2} (g)) + 3 Δ_{f} H^{o} (H_{2} O (l) - Δ_{r} H$

putting all the values,
$= 3267.0 + [6 \times (- 393.5) - 3 \times (- 285.83)]$
$= 3267.0 + [- 2361 - 857.49]$
$= 48.51 k J m o l^{- 1}$
Hence, enthalpy of formation of benzene is $48.51 k J m o l^{- 1}$

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Similar questions

Q. The reaction of 1 mol of benzene take place at

298 K

1 a t m

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l)

In this reaction

C O_{2} (g)

H_{2} O (l)

are produced and 3267.0

k J

heat is liberated. calculate the

Δ_{f} H^{o}

Δ_{f} H^{o}

C O_{2} (g)

H_{2} O (l)

- 393.5 k J m o l^{- 1}

- 285.83 k J m o l^{- 1}

Q. The combustion of one mole of benzene takes place at 298K and 1 atm after combustion

C O_{2 (g)}

H_{2} O_{J}

are produced and 3267.0 KJ of heat is liberated. Calculate the standard enthalpy of formulation

Δ H f

of benzene. Standard enthalpy of formation of

C O_{2 (g)}

H_{2} O_{J}

are -393.5Kj and -285.83Kl/Mole

Q. Benzene burns in oxygen according to the following reaction :

C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to {3 H}_{2} O (l) + {6 C O}_{2} (g)

If the standard enthalpies of formation of

C_{6} H_{6} (l)

H_{2} O (l)

{C O}_{2} (g)

are 11.7, -68.1 and -94 kcal/mole, respectively, the amount of heat that will liberate burning 780 g benzene is:

Δ_{f} H^{o}

C O_{2} (g), C O (g)

H_{2} O (l)

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k J m o l^{- 1}

respectively. The standard enthalpy change (in

k J m o l^{- 1}

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C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l)

Q. For the reaction:

4 N H_{3} (g) + 3 O_{2} (g) \to 2 N_{2} (g) + 6 H_{2} O (l)

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. Given that heat of formation of

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Heat of Reaction

Standard XII Chemistry

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