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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2016 · Chemistry · Thermodynamics (C)

For a particular reversible reaction at temperature T, Δ H and Δ S...

For a particular reversible reaction at temperature T,ΔHT, \Delta H and ΔS\Delta S were found to be both +ve. If TeT_{e} is the temperature at equilibrium, the reaction would be spontaneous when

Options

  1. ATe>TT_{e} > T
  2. BT>TeT > T_{e}
  3. CTeT_{e} is 5 times TT
  4. DT=TeT=T_{e}

Correct answer

B. T>TeT > T_{e}

Step-by-step solution

At equilibrium ΔG=0\Delta G=0 Hence, ΔG=ΔH−TeΔS=0∴ΔH=TeΔS or Te=ΔHΔS\begin{array}{l}\text{ Hence, } \Delta G=\Delta H-T_{e} \Delta S=0 \\ \therefore \Delta H=T_{e} \Delta S \text{ or } T_{e}=\frac{\Delta H}{\Delta S} \end{array}$ForaspontaneousreactionFor a spontaneous reaction\Delta Gmustbenegativewhichispossibleonlyifmust be negative which is possible only if\Delta H-T \Delta S< 0\therefore \Delta H< T \triangle SororT > \frac{\Delta H}{\Delta S}$ ; T_{e}< T
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