Question

At $\,{0^o}C\,$, ice and water are in equilibrium and enthalpy change for the process 

$\,H_2O(s) \to H_2O(l)\,$ is $\,6kJmo{l^{ - 1}}\,$ .Calculate the entropy change for the conversion of ice into liquid water.

Answer 1

Hint:For certain daily phenomena, the notion of entropy gives profound insight into the course of random transition. The principle of entropy offers a mathematical means of encoding the intuitive notion of which processes are unlikely to happen, even if the universal law of energy conservation will not be broken.
Formula used: There is only one formula that’s used here that is
$\,\Delta S = \dfrac{{\Delta H}}{T}\,$ where, $\,\Delta S\,$ is entropy change
$\,\Delta H\,$ is enthalpy change
$\,T\,$ is temperature in Kelvin

Complete step by step answer:

Let us first understand the terms mentioned in the question clearly;

At equilibrium the concentration of reactant and products remains constant but not necessarily equal. Equilibrium can only be obtained in a closed system that is the reaction should be carried out in a closed container.

Now let’s understand what’s enthalpy change. Enthalpy change is the name given to the amount of heat evolved or absorbed in a reaction carried out at constant pressure. It is given the symbol of $\,\Delta H\,$ . 

Another important term is Entropy change. This is what you need to find out, so understanding the meaning of what you need to calculate is extremely essential. Entropy is defined as the function of the state of the system, so the change in entropy of a system is determined by its initial and final states. Increasing temperature increases entropy. Changing volume changes the entropy. In simple words, entropy is a system’s thermal energy per unit temperature that is unavailable for doing useful work. It is denoted by $\,\Delta S\,$.

So, it's given that at \[\,{0^o}C\,\] ice and water are at equilibrium.

$\,{0^o}C\, = 273K\,$

Enthalpy change that is $\,\Delta H\,$$\, = \,$ $\,6kJ\,$   ($\,kJ\,$ is kilojoules)                   {given}

Therefore, required entropy change that is $\,\Delta S = \dfrac{{\Delta H}}{T}\,$

Now, we will substitute the given values,$\,\Delta S = \dfrac{{6 \times {{10}^3}J/mol}}{{273K}}\,$    ( Because $\,1kJ = 1000J\,$)

$\, = 21.9780\,$$\,J/molK\,$.

Therefore, answer is $\,21.9780J/molK\,$

Note: Entropy change in a chemical reaction is given by the sum of the entropies of the products minus the sum of the entropies of the reactants. If a reaction is exothermic, enthalpy is negative and when entropy is positive the reaction is spontaneous ( making gibbs free energy negative ) and vice versa.