The enthalpy of formation of H2O(l) is -285.77 kJ mol-1 and enthalpy of neutralisation of strong acid and strong base is -56.07 kJ mol-1, what is the enthalpy of formation of OH- ion

Question

The enthalpy of formation of H2O(l) is -285.77 kJ mol-1 and enthalpy of neutralisation of strong acid and strong base is -56.07 kJ mol-1, what is the enthalpy of formation of OH- ion (A) +229.70 kJ (B) -229.70 kJ (C) +226.70 kJ (D) -22.670 kJ

Tardigrade Admin · Accepted Answer

H+(aq)+OH-(aq)→ H2O(l);Δ H=-56.07 kJ Δ H=Δ Hf(H2O)-[Δ Hf(H+)+Δ Hf(OH-)] [∵Δ Hf(H+)=0]-56.07=-285.77-(0+x) ∴ x=-285.77+56.07=-229.70 kJ

Q. The enthalpy of formation of $H_{2} O (l) i s - 285.77 k J m o l^{- 1}$ and enthalpy of neutralisation of strong acid and strong base is $- 56.07 k J m o l^{- 1}$ , what is the enthalpy of formation of $O H^{-}$ ion

$+ 229.70 k J$

$- 229.70 k J$

$+ 226.70 k J$

$- 22.670 k J$

$H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l); Δ H = - 56.07 k J$
$Δ H = Δ H_{f} (H_{2} O) - [Δ H_{f} (H^{+}) + Δ H_{f} (O H^{-})]$
$[∵ Δ H_{f} (H^{+}) = 0] - 56.07 = - 285.77 - (0 + x)$
$∴ x = - 285.77 + 56.07 = - 229.70 k J$

Q. The enthalpy of formation of H2​O(l)is−285.77kJmol−1 and enthalpy of neutralisation of strong acid and strong base is −56.07kJmol−1, what is the enthalpy of formation of OH− ion

+229.70kJ

−229.70kJ

+226.70kJ

−22.670kJ