MHT CET · Chemistry · Thermodynamics (C)
What is the value of \(\Delta \mathrm{S}_{(\text {total })}\) for following reaction at \(300 \mathrm{~K}\) \(\mathrm{Fe}_{2} \mathrm{O}_{3(\mathrm{~s})}+3 \mathrm{CO}_{(\mathrm{g})} \longrightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+\) \(3 \mathrm{CO}_{2(\mathrm{~g})} \Delta \mathrm{H}^{\circ}=-25 \mathrm{~kJ}, \Delta \mathrm{S}^{\circ}=15 \mathrm{JK}^{-1}\)
- A \(68 \cdot 2 \mathrm{JK}^{-1}\)
- B \(98 \cdot 3 \mathrm{JK}^{-1}\)
- C \(8 \cdot 32 \mathrm{JK}^{-1}\)
- D \(-10 \cdot 0 \mathrm{JK}^{-1}\)
Answer & Solution
Correct Answer
(B) \(98 \cdot 3 \mathrm{JK}^{-1}\)
Step-by-step Solution
Detailed explanation
The reaction the exothermic,
\(
\begin{aligned}
\therefore \Delta H_{\text {sur }} &=+25 \mathrm{~kJ}=25000 \mathrm{~J} \\
\therefore \Delta S_{\text {sur }} &=\frac{\Delta H_{\text {arr }}}{T}=\frac{25000 \mathrm{~J}}{300 \mathrm{~K}}=83.33 \mathrm{JK}^{-1} \\
\Delta S_{\text {ton } 1} &=\Delta S_{\text {sys }}+\Delta S_{\text {surr }} \\
&=15 \mathrm{JK}^{-1}+83.33 \mathrm{JK}^{-1} \\
\therefore \Delta S_{\text {tow }} &=98.33 \mathrm{~J} \mathrm{~K}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
\therefore \Delta H_{\text {sur }} &=+25 \mathrm{~kJ}=25000 \mathrm{~J} \\
\therefore \Delta S_{\text {sur }} &=\frac{\Delta H_{\text {arr }}}{T}=\frac{25000 \mathrm{~J}}{300 \mathrm{~K}}=83.33 \mathrm{JK}^{-1} \\
\Delta S_{\text {ton } 1} &=\Delta S_{\text {sys }}+\Delta S_{\text {surr }} \\
&=15 \mathrm{JK}^{-1}+83.33 \mathrm{JK}^{-1} \\
\therefore \Delta S_{\text {tow }} &=98.33 \mathrm{~J} \mathrm{~K}^{-1}
\end{aligned}
\)
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