Question

Given ${\Delta _r}{S^ \circ } = - 266$ and the listed $\left[ {{S^0}_mvalues} \right]$. Calculate ${S^0}$for$F{e_3}{O_4}(s)$,
$4F{e_3}{O_4}(s)\left[ {......} \right] + {O_2}\left( g \right)\left[ {205} \right] \to 6F{e_2}{O_3}(s)\left[ {87} \right]$
A.$+ 111.1$
B.$+ 122.4$
C.$145.75$
D.$248.25$

Answer 1

Looking at the question, we get an idea that the question is related to thermodynamics and ${S^0}$represents entropy whereas ${\Delta _r}{S^0}$represents the entropy change in the reaction and it units are $J/K$. We take into consideration these units, although you will say that they are missing in the question, but while writing the answer for board examination we will have to mention them.

Formula used:
For the reaction; ${n_1}A \to {n_2}B$
$\Delta {S^0}_{reaction} =$ ${n_1}$(entropy change in product)- ${n_2}$(entropy change in reactant)
Where ${n_{1,}}{n_2}$are the number of moles of product and reactant respectively.
A is the reactant and B is the product.
${\Delta _r}{S^0}$=Entropy change in the reaction

Complete step by step answer:
The value of ${\Delta _r}{S^0}$(given in the question) confirms that the reaction is exothermic in nature. We also conclude that the exothermic reactions are spontaneous. The relation between product side and reactant side to solve the question will be as follows:
$\Delta {S^0}_{reaction} =$ ${n_1}$(entropy change in product)- ${n_2}$(entropy change in reactant)
Given,
$4F{e_3}{O_4} + {O_2} \to 6F{e_2}{O_3}$
And the net entropy change for the reaction:
$\Delta {S^0}_{reaction} = - 266J/K$
For reactants;
$\Delta {S^0}\left( {{O_2}} \right) = - 205J/K$
$\Delta {S^0}\left( {F{e_2}{O_3}} \right) = - 205J/K$
For the reaction, given in the question, we will pursue the solution as follows:
$\Delta {S^0}_{reaction} = 6\Delta {S^0}\left( {F{e_2}{O_3}} \right) - \Delta {S^0}\left( {{O_2}} \right) - 4\Delta {S^0}\left( {F{e_2}{O_3}} \right)$
Substituting the values, we obtain
$\Rightarrow - 266J/K = 6 \times 87J/K - 205J/K - 4\Delta {S^0}\left( {F{e_2}{O_3}} \right)$
Solving the above equation to get the value$\Delta {S^0}\left( {F{e_2}{O_3}} \right)$

$$\Rightarrow - 266J/K = 522J/K - 205J/K - 4\Delta {S^0}\left( {F{e_2}{O_3}} \right) \\\ \Rightarrow ( - 788 + 205)J/K = 4\Delta {S^0}\left( {F{e_2}{O_3}} \right) \\\ \Rightarrow \dfrac{{583}}{4}J/K = \Delta {S^0}\left( {F{e_2}{O_3}} \right) \\\$$

That is, $\Delta {S^0}\left( {F{e_3}{O_4}} \right) = 145.75J/K$

Therefore, option C. is the correct choice for the given question.

Additional Information:
Entropy is the measure of randomness in the system. If entropy change is negative for the reaction, this means that randomness of the system has decreased.

Note: As mentioned earlier in the solution that the exothermic reactions favor spontaneous process. The spontaneous process results in an increase of total entropy. This is the conclusion we draw from the second law of thermodynamics. Spontaneity of spontaneous reaction does not depend on the reaction rate, thus it may occur quickly or very slowly.