Question

Calculate the entropy change for the following reaction, \[{H_2}\left( g \right){\text{ }} + C{l_2}\left( g \right){\text{ }}-2HCl\left( g \right){\text{ }}at298K\]
Given that, \[{S^{ \ominus }}_{{H_{2}}} = 131J{K^{ - 1}}mo{l^{ - 1}},\;{S^ \ominus }C{l_2} = 223J{K^{ - 1}}mo{l^{ - 1\;}}and\;{S^ \ominus }HCl = 187J{K^{ - 1}}mo{l^{ - 1}}\]
A \[\Delta S = 20J{K^{ - 1}}mo{l^{ - 1}}\]
B \[\Delta S = - 20J{K^{ - 1}}mo{l^{ - 1}}\]
C\[\Delta S = 40J{K^{ - 1}}mo{l^{ - 1}}\]
D None of these

Answer 1

Hint: Entropy is defined as the measure of the thermal energy of a system per unit temperature which is not available for doing useful work. It is denoted as S. The SI unit for Entropy is Joules per Kelvin. Entropy changes at a constant temperature.

Complete Step by step answer: Entropy is a thermodynamic function that describes the number of arrangements (positions and/or energy levels) that are available to a system existing in a given state.

As per given reaction \[{H_2}\left( g \right){\text{ }} + C{l_2}\left( g \right){\text{ }}-2HCl\left( g \right){\text{ }}at298K\]
To calculate the entropy change for reactions, we simply look at the entropy of the final state minus the entropy of the initial state. The final state is the products in their standard state and the initial state is the reactants in their standard state. So for a reaction we simply sum the entropies of the products (times the number of moles in our thermochemical equation) minus the sum of the entropies of the reactants (times the number of moles in the thermochemical equation).
It is represented as \[\Delta {S^{ \ominus \;}} = \sum {S^ \ominus }\left( {Product} \right) - \sum {S^ \ominus }\left( {Reactant} \right)\Delta {S^{ \ominus \;}}\]
Now in this equation we are are putting values are as -
\[{S^{ \ominus }}_{{H_{2}}} = 131J{K^{ - 1}}mo{l^{ - 1}},\;{S^ \ominus }C{l_2} = 223J{K^{ - 1}}mo{l^{ - 1\;}}and\;{S^ \ominus }HCl = 187J{K^{ - 1}}mo{l^{ - 1}}.\]
Here , \[\Delta {S^{ \ominus \;}} = \sum {S^ \ominus }\left( {Product} \right) - \sum {S^ \ominus }\left( {Reactant} \right)\;{\text{ }}\;\]
\[\Delta {S^{ \ominus \;}} = 2 \times HCL - \left( {{H_2} + C{l_2}} \right)\]
\[\Rightarrow \Delta {S^{ \ominus \;}} = 2 \times 187 - \left( {131 + 223} \right)\]
\[\Rightarrow \Delta {S^ \ominus }\; = 374 - (131 - 223)\]
\[\Rightarrow \Delta {S^{ \ominus \;}} = 20J{K^{ - 1}}mo{l^{ - 1}}\]

 So the Option \[\left( A \right)\] is correct.

Note: Entropy is greater in malleable solids whereas it is lower in brittle and hard substances. When gas is dissolved in water the entropy decreases whereas it increases when liquid or solid is dissolved in water and With an increase in chemical complexity, the entropy also increases. As mass increases entropy increases. The relationship between enthalpy and entropy can be seen to calculate the Gibbs free energy.