Question

In conversion of lime-stone to lime, ${CaCO_3(s)->CaO(s) +CO2(g)}$ 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

• A. $1008\,K$
• B. $1200\, K$
• C. $858\,K$
• D. $1118\,K$

Answer 1

Answer: (D)

$1118\,K$

Answer 2

We know, $?G = ?H- T?S$ So, lets find the equilibrium temperature, i.e. at which $?G = 0$ $?H = T?S$ $T=\frac{179.1\times1000}{160.2}$ $=1118\,K$ So, at temperature above this, the reaction will become spontaneous. Hence, (4) is correct answer.