Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
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
Solution:
2207 Upvotes
Verified Answer
The correct answer is:
$1118 \mathrm{~K}$
$1118 \mathrm{~K}$
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.
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.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.