\[\begin{array}{*{20}{l}}{{{\rm{H}}^{\rm{ + }}}\left( {{\rm{aq}}} \right){\rm{ + O}}{{\rm{H}}^{\rm{ - }}}\left( {{\rm{aq}}} \right) \to {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)}\\{{\rm{S^\circ }}\left( {{\rm{298K}}} \right){\rm{ - 10}}{\rm{.7\;\;\; + 70}}}\end{array}\]
Standard entropy change for the above reaction is
A. 60.3\[{\rm{J}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\]
B. 80.7\[{\rm{J}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\]
C. −70\[{\rm{J}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\]
D. +10.7\[{\rm{J}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\]
Answer
299.4k+ views
Hint: Entropy is the degree of disorderness in a state of a thermodynamic system.
The entropy of one mole of a substance in the pure state at one atmospheric pressure and 298 K is called standard entropy.
Formula Used:
\[{\rm{\Delta }}{{\rm{S}}^{\rm{o}}}{\rm{ = }}\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right)} {\rm{ - }}\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)} \]
where \[{\rm{\Delta }}{{\rm{S}}^{\rm{o}}}\]= standard entropy
\[\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right)} \]= summation of entropies of products
\[\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)} \]= summation of entropies of reactants
Complete step-by-step answer: Entropy change for a reaction can be evaluated by deducting the sum of entropies of reactants from the sum of entropies of products.
The entropy of one mole of a substance in the pure state at one atmospheric pressure and 298 K is called standard entropy and is symbolized by \[{\rm{\Delta }}{{\rm{S}}^{\rm{o}}}\].
Given
\[{{\rm{S}}^{\rm{o}}}_{{\rm{(}}{{\rm{H}}_2}{\rm{O)}}}\]=70 J/K mol
\[{\rm{S}}_{\left( {{\rm{O}}{{\rm{H}}^ - }} \right)}^{\rm{o}}\]= -10.7 J/K mol
\[{\rm{S}}_{\left( {{{\rm{H}}^{\rm{ + }}}} \right)}^{\rm{o}}\]=0 J/K mol
One mole of water is produced.
For the given reaction,
\[\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right)} \]=70 J/K mol=373.4 J/K mol
One mole of hydrogen ions and hydroxyl ions are reacted.
\[\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)} \]
=(0-10.7)J/K mol
=-10.7J/K mol
Hence, the total entropy is given by
\[{\rm{\Delta }}{{\rm{S}}^{\rm{o}}}{\rm{ = }}\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right)} {\rm{ - }}\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)} \]
=[70-(-10.7)] J/K mol
=80.7 J/K mol
So, option B is correct.
Additional Information: Entropy's value relies on the quantity of matter present in a system making it an extensive property.
Its value does not depend on the path by which the system has attained a specific state.
It depends on only the initial and final state of the system.
So, it is a state function
Note: While attending to the question, the unit of entropy must be cited in each calculation step. The standard entropy of a reaction is evaluated by the deduction of the sum of entropies of reactants from the sum of entropies of products.
The entropy of one mole of a substance in the pure state at one atmospheric pressure and 298 K is called standard entropy.
Formula Used:
\[{\rm{\Delta }}{{\rm{S}}^{\rm{o}}}{\rm{ = }}\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right)} {\rm{ - }}\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)} \]
where \[{\rm{\Delta }}{{\rm{S}}^{\rm{o}}}\]= standard entropy
\[\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right)} \]= summation of entropies of products
\[\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)} \]= summation of entropies of reactants
Complete step-by-step answer: Entropy change for a reaction can be evaluated by deducting the sum of entropies of reactants from the sum of entropies of products.
The entropy of one mole of a substance in the pure state at one atmospheric pressure and 298 K is called standard entropy and is symbolized by \[{\rm{\Delta }}{{\rm{S}}^{\rm{o}}}\].
Given
\[{{\rm{S}}^{\rm{o}}}_{{\rm{(}}{{\rm{H}}_2}{\rm{O)}}}\]=70 J/K mol
\[{\rm{S}}_{\left( {{\rm{O}}{{\rm{H}}^ - }} \right)}^{\rm{o}}\]= -10.7 J/K mol
\[{\rm{S}}_{\left( {{{\rm{H}}^{\rm{ + }}}} \right)}^{\rm{o}}\]=0 J/K mol
One mole of water is produced.
For the given reaction,
\[\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right)} \]=70 J/K mol=373.4 J/K mol
One mole of hydrogen ions and hydroxyl ions are reacted.
\[\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)} \]
=(0-10.7)J/K mol
=-10.7J/K mol
Hence, the total entropy is given by
\[{\rm{\Delta }}{{\rm{S}}^{\rm{o}}}{\rm{ = }}\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{products}}} \right)} {\rm{ - }}\sum {{{\rm{S}}^{\rm{o}}}\left( {{\rm{reactants}}} \right)} \]
=[70-(-10.7)] J/K mol
=80.7 J/K mol
So, option B is correct.
Additional Information: Entropy's value relies on the quantity of matter present in a system making it an extensive property.
Its value does not depend on the path by which the system has attained a specific state.
It depends on only the initial and final state of the system.
So, it is a state function
Note: While attending to the question, the unit of entropy must be cited in each calculation step. The standard entropy of a reaction is evaluated by the deduction of the sum of entropies of reactants from the sum of entropies of products.
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