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Neutralisation Reaction

The energy re...

Question

The energy released in the neutralisation of $H_{2} S O_{4}$ and $K O H$ is $- 59.1 k J$ . Calculate the value of $Δ H$ for the reaction
$H_{2} S O_{4} + 2 K O H ⟶ K_{2} S O_{4} + 2 H_{2} O$

A

Δ H = - 436.2 k J

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B

Δ H = + 118.2 k J

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C

Δ H = - 220.2 k J

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D

Δ H = - 118.2 k J

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Solution

The correct option is D $Δ H = - 118.2 k J$
As given,
The energy released in the neutralisation of $H_{2} S O_{4}$ and $K O H$ is $- 59.1 k J$
so

$1 / 2 H_{2} S O_{4} + K O H ⟶ 1 / 2 K_{2} S O_{4} + H_{2} O Δ H = - 59.1 k J / m o l$

and
$H_{2} S O_{4} + 2 K O H ⟶ K_{2} S O_{4} + 2 H_{2} O Δ H = - 118.2 k J / m o l$

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Q. The energy released in the neutralisation of

H_{2} S O_{4}

K O H

59.1 k J

. Calculate the value of

Δ H

for the reaction

H_{2} S O_{4} + 2 K O H ⟶ K_{2} S O_{4} + 2 H_{2} O

The integral enthalpy of solution in kJ of one mole of

H_{2} S O_{4}

n

mole of water is given by:

Δ H_{s} = \frac{75.6 \times n}{n + 1.8}

Δ H

H_{2} S O_{4}

dissolved in 2 mole of

H_{2} O

Δ H

k J

for the following reaction:

C (g) + O_{2} (g) \to C O_{2} (g)

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

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

Δ H = - 282 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

Δ H = - 242 k J

S + O_{2} \to S O_{2}

Δ H = - 298.2

S O_{2} + \frac{1}{2} O_{2} \to S O_{3}

Δ H = - 98.7

S O_{3} + H_{2} O \to H_{2} S O_{4}

Δ H = - 130.2

H_{2} + \frac{1}{2} O_{2} \to H_{2} O

Δ H = - 287.3

Then the enthalpy of formation of

H_{2} S O_{4}

S + O_{2} \to S O_{2}; Δ H = - 298.2 k J

S O_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

S O_{3} + H_{2} O \to H_{2} S O_{4}; Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J

the enthalpy of formation of

H_{2} S O_{4}

298

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