Question

On the basis of the following thermochemical data: $(\Delta H_f^ \circ H_{(aq)}^ + = 0)$
${H_2}{O_{(l)}} \to {H^ + }_{(aq)} + O{H^ - }_{(aq)};\Delta H = 57.32kJ$
${H_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to {H_2}{O_{(l)}};\Delta H = - 286.20kJ$
The value of enthalpy of formation $O{H^ - }$ ion at ${25^ \circ }C$ is:
(A) -22.88kJ
(B) -228.88kJ
(C) +228.88kJ
(D) -343.52kJ

Answer 1

We can calculate enthalpy change by using the standard enthalpy of formation by following equation
${\Delta _f}H = \sum\limits_i {{a_i}\Delta {H^ \circ }_f(products) - } \sum\limits_i {{b_i}\Delta {H^ \circ }_f(reactants)} {\text{ }}......{\text{(a)}}$

Complete step by step solution:
We know that the standard enthalpy change for the formation of one mole compound from its elements is called its molar enthalpy of formation and it is shown by $\Delta {H^ \circ }_f$.
- We can use the standard enthalpy of formation value in order to calculate the enthalpy change of the reaction. For the change in enthalpy, the equation we can use is
${\Delta _f}H = \sum\limits_i {{a_i}\Delta {H^ \circ }_f(products) - } \sum\limits_i {{b_i}\Delta {H^ \circ }_f(reactants)} {\text{ }}......{\text{(a)}}$
We are given that
${H_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to {H_2}{O_{(l)}};\Delta H = - 286.20kJ$
So, for this reaction, we can say that enthalpy of formation of water molecule can be given as
$\Delta H_f^{} = \Delta H_f^ \circ ({H_2}O) - \Delta H_f^ \circ ({H_2}) - \Delta H_f^ \circ \left( {\dfrac{1}{2}} \right)({O_2}){\text{ }}...{\text{(1)}}$
We are given that $\Delta {H_f}$ is -286.20kJ and $\Delta H_f^ \circ ({H_2}){\text{ and }}\Delta H_f^ \circ ({O_2})$ will be zero. So, we can write the equation (1) as
$- 286.20kJ = \Delta H_f^ \circ ({H_2}O)$
Now, we are also provided that
${H_2}{O_{(l)}} \to {H^ + }_{(aq)} + O{H^ - }_{(aq)};\Delta H = 57.32kJ$
So, here we will also use the equation (a) in which we will take $\Delta H = 57.32kJ$ . So, we will get
$57.32kJ = \Delta H_f^ \circ (O{H^ - }) + \Delta H_f^ \circ ({H^ + }) - \Delta H_f^ \circ ({H_2}O)$
Here, we obtained that $- 286.20kJ = \Delta H_f^ \circ ({H_2}O)$ and it is given in the question that $(\Delta H_f^ \circ H_{(aq)}^ + = 0)$ So, we can write that
$57.32kJ = \Delta H_f^ \circ (O{H^ - }) + 0 - ( - 286.20kJ)$
So, we obtained that
$\Delta H_f^ \circ (O{H^ - }) = 57.32kJ - 286.20kJ = - 228.88kJ$
Thus, we found that the value of enthalpy of formation of $O{H^ - }$ at ${25^ \circ }C$ is -228.88kJ.

So, the correct answer to this question is (B).

Note: Do not forget the minus sign in the equation (a) as it may lead to errors. Note that the reference state of an element is its most stable state of aggregation at ${25^ \circ }C$ and 1 bar pressure.