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Standard XII

Chemistry

Enthalpy of Neutralisation

When 400 mL o...

Question

When $400 m L$ of $0.2 M H_{2} S O_{4}$ solution is mixed with $600 m L$ of $0.1 M N a O H$ solution, the increase in temperature of the final solution is $x \times 10^{- 2} K$
The value of x (Round off to the Nearest Integer) is
[Use : $H^{+} (a q) + O H^{-} (a q) \to H_{2} O : Δ_{γ} H = - 57.1 {kJ mol}^{- 1}$
$Specific heat of H_{2} O = 4.18 {J k}^{- 1} g^{- 1}$
$Density of H_{2} O = 1.0 {g cm}^{- 3}$
[ Assume no change in volume of solution on mixing.]

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Solution

$millimoles of H_{2} S O_{4} = 400 \times 0.2 = 80$
$meq of N a O H = 600 \times 0.1 = 60$
$H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O t = 08060 - - 50 - 30$
30 m moles of product is formed.
$H^{+} + O H^{-} \to H_{2} O Δ H = - 57.1 kJ/mol .$
$Moles of H^{+} & O H^{-} neutralised = \frac{60}{1000}$
$∴ Δ H = \frac{60}{1000} \times (- 57.1) = 3426 J/mol$
$Total volume = 1 L, Mass = 1000 g$
$m s Δ T = Δ H$
$1000 \times 4.18 \times Δ T = 3426 Δ T = 0.8196 = 81.9 \times 10^{2} K = 82 \times 10^{- 2} K$
$x = 82$

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Similar questions

At a particular temperature $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l); Δ H = - 57.1 k J$ . The approximate heat evolved when $400 m L$ of $0.2$ $H_{2} S O_{4}$ is mixed with $600 m L$ of $0.1 m$ $K O H$ solution will be:

Q. 200 mL of

0.2 M H C l

is mixed with 300 mL of

0.1 M N a O H

. The molar heat of neutralization of this reaction is

- 57.1 k J

. The increase in temperature in

^{o} C

of the system on mixing is

x \times 10^{- 2}

. The value of x is (Nearest integer)
[Given: Specific heat of water = 4.18 J

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Density of water = 1.00 g

c m^{- 3}

]
[Assume no volume change on mixing]

500 m L

0.2 M

aqueous solution of acetic acid is mixed with

500 m L

0.2 M H C I

25^{o} C

. If

6 g

N a O H

is added to the above solution, determine the final pH (assuming no change in volume on mixing.

K_{a}

of acetic acid is

1.75 \times 10^{- 5} m o l L^{- 1}

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Q. When

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of a

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solution is mixed with

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