Question

In the reaction of ${I_2}$ with water, change in free energy is.
A) Negative.
B) Positive.
C) Zero.
D) Can’t be predicted.

Answer 1

Hint: We know that Gibbs free energy is adequate to the enthalpy of the system minus the product of the temperature and entropy. The equation is given as,
\[G = H-TS\]
Where,
G is Gibbs free energy.
H is enthalpy of the reaction.
T is the temperature.
S is the entropy of the reaction.

Complete step by step answer:
We know that,
Gibbs free energy doesn’t depend upon the path in which the reaction takes place hence it is a state function. So change in Gibbs free energy is,
If the reaction is administered under constant temperature\[\left\{ {\Delta T = O} \right\}\]
\[\Delta G = \Delta H - T\Delta S\]
This equation is named the Gibbs Helmholtz equation
If \[\Delta G > 0\], then the reaction is nonspontaneous and endergonic
If\[\Delta G < 0\], then the reaction is spontaneous and exergonic.
If \[\Delta G = 0\], then the reaction is at equilibrium.
The reaction of Iodine with water is endothermic, thus the change in free energy is Positive. Therefore option B is correct.
Let us see the reaction of iodine with water,
Iodine is strongly reactive, albeit it's less extreme for iodine than for other halogens. Iodine can't be found as a component, but rather as Iodine molecules, as Iodide ions, or as iodate anion.
The reaction is,
\[{I_2}\left( l \right) + {H_2}O\left( l \right){\text{ }}\xrightarrow{{}}O{I^ - }\left( {aq} \right) + 2{H^ + }\left( {aq} \right) + {I^ - }\left( {aq} \right)\]
Iodine molecules and water molecules react to substances like hypoiodite and the reaction can move both ways of the equilibrium, counting on the pH of the solution. Iodine can also occur as \[{I_3}^ - \left( {aq} \right)\], \[HIO\left( {aq} \right)\],\[I{O^ - }(aq)\], and \[HI{O_3}\left( {aq} \right)\].
Iodine can bind to several different substances, for instance other halogens. The compounds that are formed behave differently once they are in contact with water.

Therefore option B is correct.

Note:
We must remember that the stability of a compound can be determined by the Gibb’s free energy of formation. If the Gibb’s free energy of formation of compound is negative then it is more stable and if the Gibb’s free energy of formation of compound is positive then it is less stable.
Example:
The value of Gibb’s free energy of formation of methylamine is ${\text{ + 32}}{\text{.16J}}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$. Since the value is positive methylamine is less stable than its elements at the standard conditions.