
For oxidation of iron $4F{e_{(s)}} + 3{O_{2(g)}} \to 2F{e_2}{O_{3(s)}}$ enthalpy change for reaction ${\Delta _f}{H^o} = - 1648 \times {10^3}Jmo{l^{ - 1}}$ and entropy change is $ - 549.4J{K^{ - 1}}mo{l^{ - 1}}$ at $298K$ . Calculate total entropy change for this reaction is $kJ{K^{ - 1}}mo{l^{ - 1}}$ .
Answer
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Hint:Total entropy change will be the sum of entropy of the system and entropy of surrounding. In order to calculate the entropy change for the given reaction we have to calculate the entropy of the surrounding by using the relation between enthalpy of formation and entropy of surrounding.
Complete step by step answer:
According to the question, the reaction is followed as:
$4F{e_{(s)}} + 3{O_{2(g)}} \to 2F{e_2}{O_{3(s)}}$
For spontaneous reaction, total entropy is always positive that is always greater than zero.
Here, ${\Delta _f}{H^o} = - 1648 \times {10^3}Jmo{l^{ - 1}}$ ,
As heat of formation is negative that means it is an exothermic reaction.
So, entropy of surrounding will be:
$ \Rightarrow \Delta {S_{SURROUNDING}} = \dfrac{{{\Delta _f}H}}{T}$
On substituting the values,
$\Delta {S_{SURR}} = \dfrac{{1648 \times {{10}^3}}}{{298}}$
$ \Rightarrow \Delta {S_{SURR}} = 5530Jmo{l^{ - 1}}{k^{ - 1}}$
Total entropy change will be the sum of entropy of system and entropy of surrounding.
That is:
$ \Rightarrow \Delta {S_{TOTAL}} = \Delta {S_{SYSTEM}} + \Delta {S_{SURROUNDING}}$
$\
\Delta {S_{TOTAL}} = ( - 549.4 + 5530)J{k^{ - 1}}mo{l^{ - 1}} \\
\Rightarrow \Delta {S_{TOTAL}} = 4980.6J{k^{ - 1}}mo{l^{ - 1}} \\
\ $
Hence, the total entropy change for the given reaction is $4980.6J{k^{ - 1}}mo{l^{ - 1}}$ .
Note:
$\Delta {H_f}$ represents the heat of formation, it is calculated by the difference between the enthalpy of product minus the enthalpy of reactants. It is measured in kilojoule per mole that is $kJ/mol$ . More precisely, it is the change in enthalpy for creating one mole of a compound at standard conditions. Sometimes the formation of reaction may be defined as the sum of a number of simpler reactions, does not matter if they are real or fictitious. The enthalpy of reaction can then be determined by the help of Hess's Law, this law states that the sum of the enthalpy changes for a number of individual reaction steps equals to the enthalpy change of the overall reaction. This is true because enthalpy is a state function, in state function value for an overall process depends only on the initial and final states and not on any intermediate states.
Complete step by step answer:
According to the question, the reaction is followed as:
$4F{e_{(s)}} + 3{O_{2(g)}} \to 2F{e_2}{O_{3(s)}}$
For spontaneous reaction, total entropy is always positive that is always greater than zero.
Here, ${\Delta _f}{H^o} = - 1648 \times {10^3}Jmo{l^{ - 1}}$ ,
As heat of formation is negative that means it is an exothermic reaction.
So, entropy of surrounding will be:
$ \Rightarrow \Delta {S_{SURROUNDING}} = \dfrac{{{\Delta _f}H}}{T}$
On substituting the values,
$\Delta {S_{SURR}} = \dfrac{{1648 \times {{10}^3}}}{{298}}$
$ \Rightarrow \Delta {S_{SURR}} = 5530Jmo{l^{ - 1}}{k^{ - 1}}$
Total entropy change will be the sum of entropy of system and entropy of surrounding.
That is:
$ \Rightarrow \Delta {S_{TOTAL}} = \Delta {S_{SYSTEM}} + \Delta {S_{SURROUNDING}}$
$\
\Delta {S_{TOTAL}} = ( - 549.4 + 5530)J{k^{ - 1}}mo{l^{ - 1}} \\
\Rightarrow \Delta {S_{TOTAL}} = 4980.6J{k^{ - 1}}mo{l^{ - 1}} \\
\ $
Hence, the total entropy change for the given reaction is $4980.6J{k^{ - 1}}mo{l^{ - 1}}$ .
Note:
$\Delta {H_f}$ represents the heat of formation, it is calculated by the difference between the enthalpy of product minus the enthalpy of reactants. It is measured in kilojoule per mole that is $kJ/mol$ . More precisely, it is the change in enthalpy for creating one mole of a compound at standard conditions. Sometimes the formation of reaction may be defined as the sum of a number of simpler reactions, does not matter if they are real or fictitious. The enthalpy of reaction can then be determined by the help of Hess's Law, this law states that the sum of the enthalpy changes for a number of individual reaction steps equals to the enthalpy change of the overall reaction. This is true because enthalpy is a state function, in state function value for an overall process depends only on the initial and final states and not on any intermediate states.
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