Question

For reduction of ferric oxide by hydrogen,
${{F}}{{{e}}_{{2}}}{{{O}}_{{3}}}{{(s) + 3}}{{{H}}_{{2}}}{{(g)}} \to {{2Fe(s) + 3}}{{{H}}_{{2}}}{{O(l)}}$ , ${{\Delta H}}_{{{300}}}^{{o}} = - 26.72{{kJ}}$ . The reaction was found to be too exothermic. To be convenient, it is desirable that ${{\Delta }}{{{H}}^{{o}}}$ should be at the most $- 26{{kJ}}$ . At what temperature difference is it possible?
${{{C}}_{{p}}}{{[F}}{{{e}}_{{2}}}{{{O}}_{{3}}}{{] = 105}}$ , ${{{C}}_{{p}}}{{[Fe(s)] = 25}}$ , ${{{C}}_{{p}}}{{[}}{{{H}}_{{2}}}{{O(l)] = 75}}$ , ${{{C}}_{{p}}}{{[}}{{{H}}_{{2}}}{{(g)] = 30}}$ (all are in J/mol)

Answer 1

In the above question, we are provided with an equation and ${{{C}}_{{p}}}$ values of both reactant and product , the desired ${{\Delta }}{{{H}}^{{o}}}$ value is given and asked to find out the temperature difference for which the reaction is possible. We can substitute the value of ${{\Delta }}{{{C}}_{{p}}}$ , as ${{{C}}_{{p}}}$ of all reactant and product are given, in the ${{\Delta }}{{{H}}^{{o}}}$ and hence, temperature difference can be calculated.

Formula Used
${{\Delta }}{{{H}}^{{o}}}{{ = }}\int {{{C}}_{{p}}^{{{reaction}}}{{dT}}}$
Where ${{\Delta }}{{{H}}^{{o}}}$ = enthalpy of formation
${{C}}_{{p}}^{{{reaction}}}$ = heat capacity of the reaction at constant pressure
${{dT}}$ = temperature change

Complete step by step solution:
We know that at constant pressure,
${{\Delta }}{{{H}}^{{o}}}{{ = }}\int {{{C}}_{{p}}^{{{reaction}}}{{dT}}}$
${{C}}_{{p}}^{{{reaction}}}$ can be calculated as difference between ${{{C}}_{{p}}}$ of product and ${{{C}}_{{p}}}$ of reactant.
$C_p^{{{reaction}}} = {C_p}\left( {{{product side}}} \right) - {C_p}\left( {{{reactant side}}} \right)$
For the chemical reaction,
${{F}}{{{e}}_{{2}}}{{{O}}_{{3}}}{{(s) + 3}}{{{H}}_{{2}}}{{(g)}} \to {{2Fe(s) + 3}}{{{H}}_{{2}}}{{O(l)}}$
${{C}}_{{p}}^{{{reaction}}}$ = ${{2}} \times {{{C}}_{{p}}}{{[Fe(s)]}} + 3 \times {{{C}}_{{p}}}{{[}}{{{H}}_{{2}}}{{O(l)]}} - \left( {{{3}} \times {{{C}}_{{p}}}{{[}}{{{H}}_{{2}}}{{(g)]}} + {{{C}}_{{p}}}{{[F}}{{{e}}_{{2}}}{{{O}}_{{3}}}{{]}}} \right)$
Substituting the values, we get:
${{C}}_{{p}}^{{{reaction}}} = 2 \times 25 + 3 \times 75 - (3 \times 30 + 105)$
$\Rightarrow {{C}}_{{p}}^{{{reaction}}} = 50 + 225 - (90 + 105) = 80{{J}}$
Now, substituting the value of ${{C}}_{{p}}^{{{reaction}}}$ in the formula, ${{\Delta }}{{{H}}^{{o}}}{{ = }}\int {{{C}}_{{p}}^{{{reaction}}}{{dT}}}$ , we have:
${{\Delta }}{{{H}}^{{o}}}{{ = }}\int {{{80dT}}}$
Substituting the value of ${{\Delta }}{{{H}}^{{o}}}$ ,we get:
$- 26{{kJ}} = 80{{J}}\int {{{dT}}}$
Since, $\int {{{dT}}} {{ = \Delta T}}$ , we get:
${{\Delta T}} = \dfrac{{ - 26000}}{{80}} = - 325$
$\therefore$ ${{\Delta }}{{{H}}^{{o}}}$ can be $- 26{{kJ}}$ if the temperature difference is $- 325$ kelvin or celsius.

Note:
Heat capacity of a system undergoing phase transition is infinite, as the heat is utilized in changing the state of the material instead of raising the overall temperature.
The heat capacity can usually be measured by the method which is directly implied by its definition, i.e., it starts with the object at a known uniform temperature and then a known amount of heat energy is added to it. Then we have to wait for its temperature to become uniform and then measure the change in its temperature. This method gives accurate values for many solids but it cannot provide very precise measurements for gases.