Question

The entropy (${ S }^{ o }$) of the following substances are:
${ CH }_{ 4 }(g)=186.2\quad J{ K }^{ -1 }{ mol }^{ -1 }$
${ O }_{ 2 }(g)=205.0\quad J{ K }^{ -1 }{ mol }^{ -1 }$
${ CO }_{ 2 }(g)=213.6\quad J{ K }^{ -1 }{ mol }^{ -1 }$
${ H }_{ 2 }O(l)=69.9\quad J{ K }^{ -1 }{ mol }^{ -1 }$
The entropy change ($\Delta { S }^{ o }$) for the reaction ${ CH }_{ 4 }(g)+2{ O }_{ 2 }(g)\rightarrow { CO }_{ 2 }(g)+2{ H }_{ 2 }O(l)$ is:
(a) $-312.5\quad J{ K }^{ -1 }{ mol }^{ -1 }$
(b) $-242.8\quad J{ K }^{ -1 }{ mol }^{ -1 }$
(c) $-108.1\quad J{ K }^{ -1 }{ mol }^{ -1 }$
(d) $-37.6\quad J{ K }^{ -1 }{ mol }^{ -1 }$

Answer 1

Hint: We will use Hess’s law in order to solve this question. We generally use Hess’s Law for calculating the enthalpy change of a reaction since enthalpy change is a state function but we can also apply this law to calculate the entropy change for a reaction since the entropy change is also a state function i.e. its value depends only on the initial state and the final state of the system and does not depends upon the path followed.

Complete step by step solution:
In order to solve this question, we first need to know the Hess’s Law of constant heat summation. This law was given in 1840 by G.H. Hess, a Russian chemist, on the basis of his experimental observations. It states that the total amount of heat evolved or absorbed in a reaction is the same whether the reaction takes place in one step or in a number of steps i.e. the total amount of heat change in a reaction depends only upon the nature of the initial reactants and the nature of the final products and is independent of the path or the manner by which this change is brought about. We generally use Hess’s Law for calculating the enthalpy change of a reaction since enthalpy change is a state function but we can also apply this law to calculate the entropy change for a reaction since the entropy change is also a state function i.e. its value depends only on the initial state and the final state of the system and does not depends upon the path followed.
Now let us solve the question.
It is given that:
${ CH }_{ 4 }(g)=186.2\quad J{ K }^{ -1 }{ mol }^{ -1 }$
${ O }_{ 2 }(g)=205.0\quad J{ K }^{ -1 }{ mol }^{ -1 }$
${ CO }_{ 2 }(g)=213.6\quad J{ K }^{ -1 }{ mol }^{ -1 }$
${ H }_{ 2 }O(l)=69.9\quad J{ K }^{ -1 }{ mol }^{ -1 }$
We need to find the enthalpy change for the reaction:
${ CH }_{ 4 }(g)+2{ O }_{ 2 }(g)\rightarrow { CO }_{ 2 }(g)+2{ H }_{ 2 }O(l)$
Using Hess’s Law, the entropy change will be:
${ \Delta S }^{ o }$ = entropy of products – entropy of reactants
$\Rightarrow \Delta{ S }^{ o }=[213.6+(2\times 69.9)]-[186.2+(2\times 205)]=-242.8\quad J{ K }^{ -1 }{ mol }^{ -1 }$

Hence the correct answer is (b) $-242.8\quad J{ K }^{ -1 }{ mol }^{ -1 }$

Note: Please note that for calculating the enthalpy change or entropy change for a reaction, the reaction should be balanced. The enthalpy change will be obtained by subtracting the sum of the enthalpy for the reactants multiplied by their respective stoichiometric coefficients from the sum of the enthalpy of the products multiplied by their respective coefficients.