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
B
1200 K
C
845 K
D
1118 K

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

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