Heat of neutralisation of NaOH and HCl is -57.3 kJ mol-1. The heat of ionization of water will be

Question

Heat of neutralisation of NaOH and HCl is -57.3 kJ mol-1. The heat of ionization of water will be (A) -57.3 kJ mol-1 (B) -114.6 kJ mol-1 (C) +57.3 kJ mol-1 (D) + 114.6 kJ mol-1

Tardigrade Admin · Accepted Answer

NaOH + HC1 arrow NaCl + H2O or H+ + OH- arrow H2O ; Δ H = -57.3 kJ ∴ H2O arrow H+ + OH- ; Δ H = +57.3 kJ This is according to Laplace law

Q. Heat of neutralisation of NaOH and HCl is $- 57.3 k J m o l^{- 1}$ . The heat of ionization of water will be

$- 57.3 k J m o l^{- 1}$

$- 114.6 k J m o l^{- 1}$

$+ 57.3 k J m o l^{- 1}$

$+ 114.6 k J m o l^{- 1}$

$N a O H + H C 1 \to N a Cl + H_{2} O$
or $H^{+} + O H^{-} \to H_{2} O; Δ H = - 57.3$ kJ
$∴ H_{2} O \to H^{+} + O H^{-}; Δ H = + 57.3$ kJ
This is according to Laplace law

Q. Heat of neutralisation of NaOH and HCl is −57.3kJmol−1. The heat of ionization of water will be

−57.3kJmol−1

−114.6kJmol−1

+57.3kJmol−1

+114.6kJmol−1