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

Q. When

100 m L

each of

H C l

and

N a O H

solutions are mixed,

5.71 k J

of heat was evolved. What is the molarity of two solutions? The heat of neutralisation of

H C l

57.1 k J

Join BYJU'S Learning Program

Heat of neutralization for the reaction, is 57.1kJ mol−1. The heat released when 0.25 mole of NaOH is titrated against 0.25 mole HCl will be:NaOH+HCl→NaCl+H2O

The correct option is D 14.3kJNaOH+HCl⟶NaCl+H2OHeat of neutralization for 1 mole of NaOH and HCl is 57.1KJ.∴ Heat of neutralization for 0.25 mole =0.25×57.1KJ=14.3KJ

Heat of neutralization for the reaction, is $57.1 k J$ ${m o l}^{- 1}$ . The heat released when $0.25$ mole of $N a O H$ is titrated against $0.25$ mole $H C l$ will be:
$N a O H + H C l \to N a C l + H_{2} O$

The correct option is D $14.3 k J$
$N a O H + H C l ⟶ N a C l + H_{2} O$
Heat of neutralization for 1 mole of $N a O H$ and $H C l$ is $57.1 K J$ .
$∴$ Heat of neutralization for 0.25 mole $= 0.25 \times 57.1 K J = 14.3 K J$