JEE MAIN - Chemistry (2007 - No. 8)
In conversion of lime-stone to lime,
CaCO3(s) $$\to$$ CaO(s) + CO2 (g) the vales of ∆H° and ∆S° are +179.1 kJ mol−1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ∆H° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is :
CaCO3(s) $$\to$$ CaO(s) + CO2 (g) the vales of ∆H° and ∆S° are +179.1 kJ mol−1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ∆H° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is :
1008 K
1200
845 K
1118 K
ব্যাখ্যা
$$\Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ }$$
For a spontaneous reaction $$\Delta {G^ \circ } < 0$$
or $$\,\,\,\Delta {H^ \circ } - T\Delta {S^ \circ } < 0$$
$$ \Rightarrow T > {{\Delta {H^ \circ }} \over {\Delta {S^ \circ }}}$$
$$ \Rightarrow T > {{179.3 \times {{10}^3}} \over {160.2}} > 1117.9K \approx 1118K$$
For a spontaneous reaction $$\Delta {G^ \circ } < 0$$
or $$\,\,\,\Delta {H^ \circ } - T\Delta {S^ \circ } < 0$$
$$ \Rightarrow T > {{\Delta {H^ \circ }} \over {\Delta {S^ \circ }}}$$
$$ \Rightarrow T > {{179.3 \times {{10}^3}} \over {160.2}} > 1117.9K \approx 1118K$$