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Enthalpy of Solution

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 = - 118.2 k J

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B

Δ H = + 118.2 k J

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C

Δ H = - 236.4 k J

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D

None of these

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Solution

The correct option is A $Δ H = - 118.2 k J$
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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Similar questions

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

Q. Given that:
i.

C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H = - 393.7 k J

C (d i a m o n d) ⟶ C (g r a p h i t e); Δ H = - 2.1 k J

Δ H

for burning of diamond to

C O_{2}

.

Q. Given the following thermos chemical reactions:

X O_{2} (s) + C O (g) \to X O (s) + C O_{2} (g)

Δ H = - 20.0

X_{3} O_{4} (s) + C O (g) \to 3 X O (s) + C O_{2} (g)

Δ H = + 6.0

3 X_{2} O_{3} (s) + C O (g) \to 2 X_{3} O_{4} (s) + C O_{2} (g)

Δ H = - 12.0

Δ H

in kJ for the reaction.

2 X O_{2} (s) + C O (g) \to X_{2} O_{3} (s) + C O_{2} (g)

.

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

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