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Standard XIII

Chemistry

Enthalpy of Neutralisation

At a particul...

Question

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:

3.426kJ

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13.7kJ

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5.2 kJ

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55kJ

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Solution

The correct option is A
3.426kJ

Number of equivalent $H_{2} S O_{4}$ taken

= $\frac{0.2 \times 2 \times 400}{1000} = 0.16$

Number of equivalent $K O H$ added

= $\frac{600 \times 0.1}{1000} = 0.06$

Number of equivalent of acid and bases which neutralized each other = $0.06$

$∴$ heat evolved = $0.06 \times 57.1 k J$ = $3.426 k J$

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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. When

400 m L

0.2 M H_{2} S O_{4}

solution is mixed with

600 m L

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.]

Q. 100 mL of a 0.2 M

H_{2} S O_{4}

solution, 200 mL of a 0.1 M

H N O_{3}

solution and 300 mL of a 0.2 M NaOH solution are mixed. The resultant solution will be

Q. When 50

{c m}^{3}

of 0.2 N

H_{2} {S O}_{4}

is mixed with 50

{c m}^{3}

of 1N KOH, the heat liberated is: (Given

H^{+} (a q) + {O H}^{-} (a q) \to H_{2} O (I); △_{n e u t} H = - 57.3 k J

)

20

mL of

0.2 M N a O H (a q)

solution is mixed with

35

mL of

0.1 M N a O H (a q)

solution and the resultant solution is diluted to

100

mL.

40

mL of this diluted solution reacted with

10 %

impure sample of oxalic acid

(H_{2} C_{2} O_{4})

. The weight of impure sample is :

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At a particular temperature H+(aq) + OH−(aq) → H2O(l); ΔH = −57.1kJ. The approximate heat evolved when 400 mL of 0.2 H2SO4 is mixed with 600 mL of 0.1 m KOH solution will be:

The correct option is A 3.426kJ Number of equivalent H2SO4 taken = 0.2 × 2 × 4001000 = 0.16 Number of equivalent KOH added = 600 × 0.11000 = 0.06 Number of equivalent of acid and bases which neutralized each other = 0.06 ∴ heat evolved = 0.06 × 57.1 kJ = 3.426 kJ

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: