The enthalpy of neutralization of HCl and NaOH is −57 kJ mol&#x

Question - The enthalpy of neutralization of

H C l

and

N a O H

- 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

N a O H

is equal to

Options:

A . 57 × 34

B . 57 × 0.5

C . 57

D . 57 × 0.25

Answer: Option A
:
A

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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