Question

If $900 \ \dfrac{J}{g}$ of heat is exchanged at boiling point of water, then what is the increase in entropy?
A. $43.4 \ \dfrac{J}{K}mol$
B. $87.2 \ \dfrac{J}{K}mol$
C. $900 \ \dfrac{J}{K}mol$
D. Zero

Answer 1

Hint: Enthalpy change (\[{\rm{\Delta H}}\]) is defined as the total heat content of the system at constant pressure. Entropy is the extent of disorder of randomness in a system. Entropy change (\[{\rm{\Delta S}}\]) of a substance measures the disorder or randomness in a system.

Formula used Entropy change (\[{\rm{\Delta S}}\]) can be calculated by using the relationship as shown below.
\[{\rm{\Delta S = }}\dfrac{{{\rm{\Delta H}}}}{{\rm{T}}}\]
where, \[{\rm{\Delta S = }}\] change in entropy
\[{\rm{\Delta H = }}\] change in enthalpy (or amount of heat supplied)
\[{\rm{T = }}\] temperature

Complete Step by Step Solution:
Molar mass of water is known to be \[{\rm{18}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]
Heat exchanged at boiling point of water= $$900 \ \dfrac{J}{g}$$

Convert the heat exchanged into units of $ \dfrac{J}{mol}$ by using the molar mass of water as:
$900 \ \dfrac{J}{g} \times 18 \ \dfrac{g}{mol} \\ = \ 16200 \ \dfrac{J}{mol}$

Again, boiling point of water is known to be 100 ^oC
So, temperature \[{\rm{(T) = 10}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\]
Convert the temperature from degree Celsius to Kelvin by using the relationship, \[{\rm{K}}{{\rm{ = }}^{\rm{0}}}{\rm{C}} + 273\]as:
\[{\rm{T = 10}}{{\rm{0}}^{\rm{0}}}{\rm{C = (100 + 273)K = 373K}}\]

Now, find the increase in entropy as shown below:
$\Delta S = \dfrac{\Delta H}{T}\\
= \dfrac{16200 \ \dfrac{J}{mol}}{373 \ K}\\
= 43.4 \ \dfrac{J}{K}mol$
Hence, the increase in entropy is found to be $43.4 \ \dfrac{J}{K}mol$
Therefore, option A is correct.

Note:Entropy is a state function like enthalpy and internal energy. So, it depends upon the final and initial states of a system. Thus, entropy change can be written as; \[{\rm{\Delta S}} = {{\rm{S}}_{{\rm{final state}}}} - {{\rm{S}}_{{\rm{initial state}}}}\]when a system undergoes a change from initial state to final state. For any chemical process, \[{\rm{\Delta S}} = {{\rm{S}}_{{\rm{(products)}}}} - {{\rm{S}}_{{\rm{(reactants)}}}}\]. For a reversible process, \[{\rm{\Delta S}} = \dfrac{{{{\rm{q}}_{{\rm{rev}}}}}}{{\rm{T}}}\]at equilibrium; where \[{{\rm{q}}_{{\rm{rev}}}}\] is the amount of heat supplied at temperature \[{\rm{T}}\]in a reversible process. Since \[{\rm{\Delta S}} = \dfrac{{{{\rm{q}}_{{\rm{rev}}}}}}{{\rm{T}}}\], therefore unit of entropy in S.I. units is Joule per Kelvin, $\dfrac{J}{K}mol$ (expressed as entropy unit E.U.). It is also expressed in calories per degree, \[{\rm{cal}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}\](expresses as entropy unit, eu) .