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

11.46 kJ

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57.3 kJ

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0.563 kJ

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0.573 kJ

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Solution

The correct option is D 11.46 kJ
Given $50 c m^{3}$ of $0.2 N$ , $H_{2} S O_{4}$ $50 c m^{3}$ of $1 N$ $K O H$
so only $0.2 N$ of $K O H$ will react
$H_{2} S O_{4} + 2 K O H \to K_{2} S O_{4} + 2 H_{2} O$
$Δ H_{n e t}$ = $2 \times (50 \times 0.2) \times 57.3 K J$
$= 20 \times 57.3 K J$
$= 11.46 K J$

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Similar questions

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 ?

Q. When

50 c m^{3}

0.2 N H_{2} S O_{4}

is mixed with

50 c m^{3}

1 N K O H

, the heat liberated is:

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 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous silution, 108 kJ of heat is liberated, then the enthalpy of ionisation of the oxalic acid is (Given,

H^{+}

(aq)+

O H^{-}

(aq)

\to

H_{2} O

(I);

△

_neut H = -57.3 kJ)

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When 50 cm3 of 0.2 N H2SO4 is mixed with 50 cm3 of 1N KOH, the heat liberated is: (Given H+(aq)+OH−(aq)→H2O(I);△neutH=−57.3kJ)

The correct option is D 11.46 kJGiven 50cm3 of 0.2N , H2SO4 50cm3 of 1N KOH so only 0.2N of KOH will reactH2SO4+2KOH→K2SO4+2H2OΔHnet = 2×(50×0.2)×57.3KJ =20×57.3KJ =11.46KJ

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