
Calculate the entropy change for the following reversible process:
$\alpha - Tin \rightleftarrows \beta - Tin$ at ${13^ \circ }C$ $\left( {\Delta {H_{trans}} = 2090J/mol} \right)$
Answer
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Hint: The given reaction is a reversible reaction. The change in enthalpy is in the transition state. So by using free energy relationships we will find the entropy change of the reversible process as there is no other relationship between entropy and enthalpy.
Complete step by step answer:
Change in entropy is dependent on the temperature, volume, pressure and the number of moles that is present in a system. This is called the state function.
Change in enthalpy is defined as the amount of energy released or absorbed at constant pressure.
The relationship between change in enthalpy and change in entropy is expressed in terms of free energy:
$G = H - TS$ .
Where, $G = $ free energy, $H = $ Enthalpy, $T = $ temperature, $S = $ Entropy.
But when there is change in free energy the equation is given as follows:
$\Delta G = \Delta H - T\Delta S$ …1
Where, $\Delta G = $ change in free energy, $\Delta H = $ change in Enthalpy, $T = $ temperature, $\Delta S = $ change in Entropy.
At equilibrium change in free energy $\left( {\Delta G} \right) = 0$ ,
Therefore, $T\Delta S = \Delta H$
Rearranging the equation we get,
$\Delta S = \dfrac{{\Delta H}}{T}$ .
This is the equation that shows the relationship between enthalpy and entropy.
Using this equation we will find the entropy change in the following reaction:
$\alpha - Tin \rightleftarrows \beta - Tin$at${13^ \circ }C$$\left( {\Delta {H_{trans}} = 2090J/mol} \right)$
Given data:
$\Delta {H_{trans}} = 2090J/mol$
$T = {13^ \circ }C$
$\therefore T = 273 + 13$
$\therefore T = 290K$
$\Delta S = \dfrac{{\Delta H}}{T}$
Substituting the values of entropy change and temperature in above equation we get:
$ \Rightarrow \Delta S = \dfrac{{2090}}{{290}}$
$ \Rightarrow \Delta S = 7.20J/mol{K^{ - 1}}$
Therefore the entropy change for the reversible process is $7.20J/mol{K^{ - 1}}$ .
So, the correct answer is “Option A”.
Note: The value of an entropy is dependent on the amount of substance that is present in the system. That is why it is called the extensive property. It is very difficult to calculate entropy so we calculate the change in entropy by accompanying the change in the given system.
Complete step by step answer:
Change in entropy is dependent on the temperature, volume, pressure and the number of moles that is present in a system. This is called the state function.
Change in enthalpy is defined as the amount of energy released or absorbed at constant pressure.
The relationship between change in enthalpy and change in entropy is expressed in terms of free energy:
$G = H - TS$ .
Where, $G = $ free energy, $H = $ Enthalpy, $T = $ temperature, $S = $ Entropy.
But when there is change in free energy the equation is given as follows:
$\Delta G = \Delta H - T\Delta S$ …1
Where, $\Delta G = $ change in free energy, $\Delta H = $ change in Enthalpy, $T = $ temperature, $\Delta S = $ change in Entropy.
At equilibrium change in free energy $\left( {\Delta G} \right) = 0$ ,
Therefore, $T\Delta S = \Delta H$
Rearranging the equation we get,
$\Delta S = \dfrac{{\Delta H}}{T}$ .
This is the equation that shows the relationship between enthalpy and entropy.
Using this equation we will find the entropy change in the following reaction:
$\alpha - Tin \rightleftarrows \beta - Tin$at${13^ \circ }C$$\left( {\Delta {H_{trans}} = 2090J/mol} \right)$
Given data:
$\Delta {H_{trans}} = 2090J/mol$
$T = {13^ \circ }C$
$\therefore T = 273 + 13$
$\therefore T = 290K$
$\Delta S = \dfrac{{\Delta H}}{T}$
Substituting the values of entropy change and temperature in above equation we get:
$ \Rightarrow \Delta S = \dfrac{{2090}}{{290}}$
$ \Rightarrow \Delta S = 7.20J/mol{K^{ - 1}}$
Therefore the entropy change for the reversible process is $7.20J/mol{K^{ - 1}}$ .
So, the correct answer is “Option A”.
Note: The value of an entropy is dependent on the amount of substance that is present in the system. That is why it is called the extensive property. It is very difficult to calculate entropy so we calculate the change in entropy by accompanying the change in the given system.
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