Question

Calculate the entropy change [ J$/$(mol.K) ] of the given reaction. The molar entropies [ J$/$(K.mol )] are given in brackets after each substance:
\[2PbS\left( s \right)\left[ {91.2} \right] + 3O2\left( g \right)\left[ {205.1} \right] \to 2PbO\left( s \right)\left[ {66.5} \right] + 2SO2\left( g \right)\left[ {248.2} \right]\]
A.$ - 113.5$
B.$ - 168.3$
C.$ + 72.5$
D.$ - 149.2$

Answer 1

Hint: We calculate the entropy change for reactions by simply looking at the entropy of the final state subtracted from 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.

Complete step by step answer:
The formula for entropy change is as follows:
\[\Delta S = \] (Sum of entropy of product)$ - $(Sum of entropy of reactant)
We are given values of entropies of reactants and products in the question above.
Now substituting the entropy values in the formula for entropy change gives us the following:
\[\Delta S = \]\[\left\{ {\left( {2 \times 66.5} \right) + \left( {2 \times 248.2} \right)} \right\} - \left\{ {\left( {2 \times 91.2} \right) + \left( {3 \times 205.1} \right)} \right\}\]
\[\Delta S = \]−\[168.3molKJ\]
A negative change in entropy indicates that the disorder of an isolated system has been decreased. For example, the reaction by which liquid water freezes into ice expresses an isolated decrease in entropy because liquid particles are more disordered than solid particles.

So the correct answer is option B.

Note: Entropy is given the symbol $S$, and standard entropy (measured at $298K$ and a pressure of $1bar$) is given the symbol ${S^ \circ }$. The units of entropy are $JK/mol$.
Entropy $(S)$ defines the degree of randomness or disorder in a system. The property that matter possesses in terms of the way energy is dispersed is known as its entropy. In a spontaneous process, the entropy of the energy increases. The entropy increases with mass, melting, vaporization or sublimation, chemical complexity, etc.

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible.It also states that the changes in entropy in the universe can never be negative.
$\Delta {S_{univ}} = \Delta {S_{sys}} + \Delta {S_{surr}}> 0$