Question

What will be the entropy change in surrounding when $1{\text{ }}mol$ of ${H_2}O$ (l) is formed under standard condition: $\Delta H_f^ \circ = - 286kJ/mol$
(A) $400J/K$
(B) $0$
(C) $959.7J/K$
(D) $37J/K$

Answer 1

Hint : To find out the entropy change during the water formation we will write the water formation equation from $Hydrogen$ and $Oxygen$ . This will give us the number of moles of ${H_2}O$ formed. Also remember the amount of energy released is equal to the amount of heat absorbed by the surrounding.

Complete Step By Step Answer:
Entropy is associated with a state of disorder, randomness, or uncertainty. Entropy refers to the measure of a system's thermal energy per unit temperature that is unavailable for doing work. Since work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder or randomness of a system.
We can find examples of entropy in our everyday life. For example smoke coming from burning of a wood, melting of ice or dissolving salt in our lemonade are all examples of entropy.
According to the second law of thermodynamics, the total entropy of a system either increases or remains constant, it never decreases. Entropy is zero in a reversible process and increases in an irreversible process.
Now that we are familiar with Entropy , we can define the Entropy change which is the change in randomness or disorder in a thermodynamic system when heat or enthalpy is converted into work.
Entropy increases with increase in the temperature of the system.
To determine the change in entropy or entropy change we use the formula
$\Delta S = \dfrac{Q}{T}$
Where
$Q$ is the heat transfer for the internally reversible system
$T$ is the temperature of the system
As mentioned in the hint lets find out the number of moles of water formed
${H_2} + \dfrac{1}{2}{O_2} \to {H_2}{O_{(l)}}$
It is clear from the above equation that $1{\text{ }}mol$ of water is formed. This further translates that at $298K$ when $1{\text{ }}mol$ of ${H_2}{O_{(l)}}$ is formed, $286KJ$ of heat is released.
Then,
$Q$ will be $286KJ/mol$ and $T$ will be $298K$
Putting these values in the Entropy change equation
$\Delta S = \dfrac{Q}{T}$
$\Delta S = \dfrac{{286}}{{298}}$
$\Delta S = 0.9597KJ/mol$
$\Delta S = 959.7J/mol$
Therefore our answer is option number C.

Note :
The formation of water is an exothermic process therefore the heat will be released or given out by the system to the surrounding. Also note that one $1KJ$ energy is equal to $1000J$ of the energy, this will help in conversion of energy units.