ThermochemistryHard

Question

The enthalpy of neutralization of a strong acid by a strong base is −57.32 kJ mol–1. The enthalpy of formation of water is −285.84 kJ mol–1. The enthalpy of formation of aqueous hydroxyl ion is

Options

A.+228.52 kJ/mol
B.−114.26 kJ/mol
C.−228.52 kJ/mol
D.+114.2 kJ/mol

Solution

$H^{+}(aq) + OH^{-}(aq) \rightarrow H_{2}O(l)$

$\Delta H = \Delta_{f}H_{H_{2}O(l)} - \left\lbrack \Delta_{f}H_{H^{+}(aq)} + \Delta_{f}H_{OH^{-}(aq)} \right\rbrack $$${\text{Or, } - 57.32 = ( - 285.84) - \left\lbrack 0 + \Delta_{f}H_{OH^{-}(aq)} \right\rbrack }{\therefore\Delta_{f}H_{OH^{-}(aq)} = - 228.52\text{ kJ/mol}}$$

Create a free account to view solution

View Solution Free
Topic: Thermochemistry·Practice all Thermochemistry questions

More Thermochemistry Questions