Question

In conversion of limestone to lime, $CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)$, the values of $\Delta {{H}^{\circ }}$ and $\Delta {{S}^{\circ }}$ are +179.1 KJ/mol and 160.2 J/K respectively at $298K$ and 1 bar. Above what temperature, conversion of limestone to lime will be spontaneous (assuming $\Delta {{H}^{\circ }}$ and $\Delta {{S}^{\circ }}$ do not change with temperature)?
(a)- 1118 K
(b)- 1008 K
(c)- 1200 K
(d)- 845 K

Answer 1

Hint: This question can be solved by the formula \[~\Delta {{G}^{\circ }}=\Delta {{H}^{\circ }}-T\Delta {{S}^{\circ }}\], where $\Delta {{G}^{\circ }}$ is the standard free energy change, $\Delta {{H}^{\circ }}$is the standard enthalpy change, T is the temperature, and $\Delta {{S}^{\circ }}$ is the change in standard entropy. For a system or reaction to occur spontaneously the change in free energy must be less than 0 or must be negative.

Complete step by step answer:
We know the formula for the change in free energy of the reaction is:
\[~\Delta {{G}^{\circ }}=\Delta {{H}^{\circ }}-T\Delta {{S}^{\circ }}\]
Where $\Delta {{G}^{\circ }}$ is the standard free energy change, $\Delta {{H}^{\circ }}$is the standard enthalpy change, T is the temperature, and $\Delta {{S}^{\circ }}$ is the change in standard entropy. For a system or reaction to occur spontaneously the change in free energy must be less than 0 or must be negative.
So we can write:
$\Delta {{G}^{\circ }}$ < 0
Now putting this value in the equation:
\[\Delta {{H}^{\circ }}-T\Delta {{S}^{\circ }}\text{ 0}\]
We can also write:
\[~\dfrac{\Delta {{H}^{\circ }}}{\Delta {{S}^{\circ }}}\text{ T}\]
So in the question the values of $\Delta {{H}^{\circ }}$ and $\Delta {{S}^{\circ }}$ are +179.1 KJ/mol and 160.2 J/K respectively. $\Delta {{H}^{\circ }}$is in KJ and $\Delta {{S}^{\circ }}$is in J, so the value of $\Delta {{H}^{\circ }}$will be multiplied with 1000.
Now putting the values,
\[~\dfrac{179.1\text{ x 1000}}{160.2}\text{ T}\]
$1117.9\text{ T}$
Or we can write:
$1118\text{ T}$
So must be around 1118 K or higher than 1118 K for the reaction to occur spontaneously.

Therefore, the correct answer is option (a)- 1118 K.

Note: If the reaction must be in equilibrium the change in free energy must be equal to 0. If the reaction must be spontaneous the change in free energy must be less than zero and if the reaction must be non-spontaneous the change in free energy must be greater than zero.