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Question: In conversion of limestone to lime, \(CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)\), the values of \(\De...

In conversion of limestone to lime, CaCO3(s)→CaO(s)+CO2(g)CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g), the values of ΔH∘\Delta {{H}^{\circ }} and ΔS∘\Delta {{S}^{\circ }} are +179.1 KJ/mol and 160.2 J/K respectively at 298K298K and 1 bar. Above what temperature, conversion of limestone to lime will be spontaneous (assuming ΔH∘\Delta {{H}^{\circ }} and ΔS∘\Delta {{S}^{\circ }} do not change with temperature)?
(a)- 1118 K
(b)- 1008 K
(c)- 1200 K
(d)- 845 K

Explanation

Solution

This question can be solved by the formula  ΔG∘=ΔH∘−TΔS∘~\Delta {{G}^{\circ }}=\Delta {{H}^{\circ }}-T\Delta {{S}^{\circ }}, where ΔG∘\Delta {{G}^{\circ }} is the standard free energy change, ΔH∘\Delta {{H}^{\circ }}is the standard enthalpy change, T is the temperature, and ΔS∘\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:
 ΔG∘=ΔH∘−TΔS∘~\Delta {{G}^{\circ }}=\Delta {{H}^{\circ }}-T\Delta {{S}^{\circ }}
Where ΔG∘\Delta {{G}^{\circ }} is the standard free energy change, ΔH∘\Delta {{H}^{\circ }}is the standard enthalpy change, T is the temperature, and ΔS∘\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:
ΔG∘\Delta {{G}^{\circ }} < 0
Now putting this value in the equation:
ΔH∘−TΔS∘ 0\Delta {{H}^{\circ }}-T\Delta {{S}^{\circ }}\text{ 0}
We can also write:
 ΔH∘ΔS∘ T~\dfrac{\Delta {{H}^{\circ }}}{\Delta {{S}^{\circ }}}\text{ T}
So in the question the values of ΔH∘\Delta {{H}^{\circ }} and ΔS∘\Delta {{S}^{\circ }} are +179.1 KJ/mol and 160.2 J/K respectively. ΔH∘\Delta {{H}^{\circ }}is in KJ and ΔS∘\Delta {{S}^{\circ }}is in J, so the value of ΔH∘\Delta {{H}^{\circ }}will be multiplied with 1000.
Now putting the values,
 179.1 x 1000160.2 T~\dfrac{179.1\text{ x 1000}}{160.2}\text{ T}
1117.9 T1117.9\text{ T}
Or we can write:
1118 T1118\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.