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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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In conversion of limestone to lime, \[\] $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$ \[\] the values of $ \Delta H^{\circ}\: and\: \Delta S^{\circ}$ are 179 kJ/mol and 160.2 J/K respectively at 298 K and 1 bar. Assuming that $ \Delta H^{\circ} \: and \: \Delta S^{\circ}$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

$\begin {array} {1 1} (A)\;1008\: K & \quad (B)\;1200\: K \\ (C)\;845\: K & \quad (D)\;1118\: K \end {array}$

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$ \Delta S = \large\frac{ \Delta H}{T}$
$T = \large\frac{ \Delta s}{ \Delta H}$
$=\large\frac{ 179.1 \times 10^3}{160.2}$$ = 1117.97 \: K = 1118\: K$
Ans : (D)
answered Mar 18, 2014 by thanvigandhi_1
 

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