MHT CET · Chemistry · Chemical Kinetics
For a reaction \(\mathrm{A}+\mathrm{B} \rightarrow\) products \(\Delta \mathrm{H}\) is \(-84.2 \mathrm{~kJ}\) and \(\Delta \mathrm{S}\) is \(-200 \mathrm{~J} \mathrm{~K}^{-1}\). Calculate the highest value of temperature so that the reaction will proceed in forward direction.
- A \(421 \mathrm{~K}\)
- B \(237 \mathrm{~K}\)
- C \(168 \mathrm{~K}\)
- D \(273 \mathrm{~K}\)
Answer & Solution
Correct Answer
(A) \(421 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
\mathrm{T} & =\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} \\
\therefore \quad \mathrm{T} & =\frac{-84200 \mathrm{~J}}{-200 \mathrm{~J} \mathrm{~K}^{-1}}=421 \mathrm{~K}
\end{aligned}\)
Since \(\Delta H\) and \(\Delta S\) are negative, the reaction is spontaneous at low temperatures. Therefore, the highest temperature is \(421 \mathrm{~K}\) and the reaction will proceed in forward direction spontaneously below \(421 \mathrm{~K}\).
\mathrm{T} & =\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} \\
\therefore \quad \mathrm{T} & =\frac{-84200 \mathrm{~J}}{-200 \mathrm{~J} \mathrm{~K}^{-1}}=421 \mathrm{~K}
\end{aligned}\)
Since \(\Delta H\) and \(\Delta S\) are negative, the reaction is spontaneous at low temperatures. Therefore, the highest temperature is \(421 \mathrm{~K}\) and the reaction will proceed in forward direction spontaneously below \(421 \mathrm{~K}\).
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