CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In conversion of lime-stone to lime, $$CaCO_{3}(s)\rightarrow CaO(s) + CO_{2}(g)$$, the values of $$\Delta^0H$$ and $$\Delta S^0 $$ are $$+179.1\ kJ mol^{-1}$$ and $$160.2\ J/Kmol$$ respectively at $$298\ K$$ and $$1$$ bar. Assuming that $$\Delta H^o$$ and $$\Delta S^o$$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is:


A
1008 K
loader
B
1200 K
loader
C
845 K
loader
D
1118 K
loader

Solution

The correct option is D $$1118\ K$$
Gibbs free energy, G is the measure of the spontaneity of a chemical process. The negative G indicates that a reaction is spontaneous under the given conditions.  

We know, $$\Delta G=\Delta H-T\Delta S$$, if $$G<0$$ then, $$\Delta H< T\Delta S$$.

Therefore, $$T=\dfrac{\Delta H}{ \Delta S} = \dfrac {(179.1 \times 1000)}{160.2} = 1118 K$$

Thus, above 1118K, $$\Delta G$$ will be negative and the reaction will be spontaneous.

Hence, the correct option is $$\text{D}$$

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image