Question

In the conversion of limestone to lime,
$\text{CaC}{{\text{O}}_{3}}\text{(s) }\to \text{ CaO(s) + C}{{\text{O}}_{2}}\text{(g)}$
the values of $\Delta H{}^\circ$ and $\Delta S{}^\circ$are $179.1\operatorname{kJ}$and $160.2\operatorname{kJ}{{\operatorname{K}}^{-1}}$respectively at $298\operatorname{K}$ and $1\operatorname{bar}$ pressure. Assuming that, $\Delta H{}^\circ$ and $\Delta S{}^\circ$ do not change with temperature; temperature above which conversion of limestone to lime will be spontaneous is:
(A) $1118\operatorname{K}$
(B) $1008\operatorname{K}$
(C) $1200\operatorname{K}$
(D) $845\operatorname{K}$

Answer 1

First take into consideration all the given quantities and change them to the standard units if required. Use the equation where free energy change is related to change in enthalpy and entropy at a particular temperature.

Complete Solution :
The given quantities are change in enthalpy ($\Delta H{}^\circ$) and change in entropy ($\Delta S{}^\circ$) at $298\operatorname{K}$ and $1\operatorname{bar}$ pressure. We all know the following equation which equates the change in Gibbs free energy to the above-mentioned quantities:
$\Delta \text{G = }\Delta \text{H }-\text{ T}\Delta \text{S}$
- We have to find out the temperature above which the given reaction will be spontaneous. For a spontaneous reaction change in free energy should be negative, that implies-
$\Delta G\,<\,0$
Substituting the value of $\Delta G$from the equation mentioned above we get:
$\Delta H\,-\,T\Delta S\,<\,0$
Putting the respective values as given in the question, we get-
$T>\dfrac{179.1}{160.2}\,K$
$=T>1.1179K$
Solving the above equation, we get “T” should be greater than $1117.9\approx 1118\operatorname{K}$.
So, the correct answer is “Option A”.

Note: The free energy change is equal to zero only when the reaction is in equilibrium. A reaction at equilibrium is also spontaneous in nature. So, in the above equation it would not be wrong to write something as:
$\Delta H-T\Delta S\le 0$
$=\dfrac{\Delta H}{\Delta S}\le T$
- The question has some extra data such as the temperature and pressure at the time of recording the experiment. This data is not required to solve the problem. Such things are done sometimes to confuse students, so caution is needed in these cases.