Q40 / 120 · JEE Main 2007 · Chemistry

In conversion of lime-stone to lime, $\mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$ the vales of $\Delta \mathrm{H}^{\circ}$ and $\Delta \mathrm{S}^{\circ}$ are $+179.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $160.2 \mathrm{~J} / \mathrm{K}$ respectively at $298 \mathrm{~K}$ and $1\mathrm{~bar}$. Assuming that $\Delta \mathrm{H}^{\circ}$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

  • A
    $1008 \mathrm{~K}$
  • B
    $1200$
  • C
    $845 \mathrm{~K}$
  • Correct answer
    D
    $1118 \mathrm{~K}$

Solution

We know, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ So, lets find the equilibrium temperature, i.e. at which $\begin{aligned} & \Delta \mathrm{G}=0 \\ & \Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} \\ & \mathrm{T}=\frac{179.1 \times 1000}{160.2} \\ & =1118 \mathrm{~K} \end{aligned}$ So, at temperature above this, the reaction will become spontaneous. Hence, (D) is correct answer.

Hence, the correct answer is option: 'D'.

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