The enthalpy of neutralization of $H C l$ and $N a O H$ is $- 57 k J m o l^{- 1}$ . The heat evolved at constant pressure (in kJ) when $0.5 m o l e$ of H2SO4 react with $0.75 m o l e$ of $N a O H$ is equal to

$57 \times \frac{3}{4}$

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$57 \times 0.5$

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$57$

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$57 \times 0.25$

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Solution

The correct option is A
$57 \times \frac{3}{4}$

$H^{-} + O H^{-} \to H_{2} O + 57 k J m o l^{- 1}$

$n_{H^{+}} = 2 \times n_{H_{2} S O_{4}} = 2 \times 0.5 = 1.0$

$n_{O H} = n_{N a O H} = 0.75$

Heat evolved = $0.75 \times 57 = \frac{3}{4} \times 57 k J$

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The enthalpy of neutralization of $H C l$ and $N a O H$ is $- 57 k J m o l^{- 1}$ . The heat evolved at constant pressure (in kJ) when $0.5 m o l e$ of H2SO4 react with $0.75 m o l e$ of $N a O H$ is equal to

The heat of neutrailisation of NaOH and HCL is 57.3 KJ/mole.What will be the amount of heat evolved when 2mole of NaOH and HCL are reacted

A: Enthalpy of neutralization of 1 equivalent each of HCI and

H_{2} S O_{4}

with NaOH is same.

R: Enthalpy of neutralization is always the heat evolved when 1 mole acid is neutralized by a base

The heat of neutralization of a strong base and a strong acid is 57 kJ. The heat released, when 0.5 mole of $H N O_{3}$ solution is added to 0.20 moles of NaOH solution, is:

Q. The enthalpy of neutralization of 1M HCI by 1M NaOH is -57 kJ/mole. The enthalpy of formation of water is -285 kJ/mole. The enthalpy of formation of

O H^{-}

ion is :

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The enthalpy of neutralization of HCl and NaOH is −57 kJ mol−1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4 react with 0.75 mole of NaOH is equal to

The correct option is A 57 × 34 H− + OH− → H2O + 57 kJ mol−1 nH+ = 2 × nH2SO4 = 2 × 0.5 = 1.0 nOH = nNaOH = 0.75 Heat evolved = 0.75 × 57 = 34 × 57 kJ

The enthalpy of neutralization of $H C l$ and $N a O H$ is $- 57 k J m o l^{- 1}$ . The heat evolved at constant pressure (in kJ) when $0.5 m o l e$ of H2SO4 react with $0.75 m o l e$ of $N a O H$ is equal to

The correct option is A
$57 \times \frac{3}{4}$

$H^{-} + O H^{-} \to H_{2} O + 57 k J m o l^{- 1}$

$n_{H^{+}} = 2 \times n_{H_{2} S O_{4}} = 2 \times 0.5 = 1.0$

$n_{O H} = n_{N a O H} = 0.75$

Heat evolved = $0.75 \times 57 = \frac{3}{4} \times 57 k J$