The reaction of 1 mole of benzene take place at 298 K and 1 atm asC 6 H 6 l -15-2 O 2 g - 6 CO 2 g -3 H 2 O l in this reaction CO 2 g and H 2 O l are produced and 3267-0 kJ heat is liberated- calculate the - f H 0 for benzene- - f H 0 for CO 2 g and H 2 O l are -393-5 kJ mol -1 and -285-83 kJ mol -1 respectively-A- 68-51 Jmol -1B- 48-51 kJmol -1C- -70-51 Jmol -1D- 55-51 kJmol -1

The correct option is B 48.51 kJmol^ 1 nbsp;Given, Δ_fH^oCO_2(g)= 393.5 kJmol^ 1, Δ_fH^oH_2O(l)= 285.83 kJmol^ 1 Reaction is: C_6H_6(l)+152O_2(g)→6CO ...

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Standard XII

Chemistry

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.

68.51 J m o l^{- 1}

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- 70.51 J m o l^{- 1}

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55.51 k J m o l^{- 1}

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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

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}

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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.

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

C O_{2 (g)}

and

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

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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)

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{C O}_{2} (g)

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Q. The

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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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Heat of Reaction

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