Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIFree Energy and EquilibriumMedium2 minPYQ_2019

ChemistryMediummatching list

A process has $Δ H = 200 J m o l^{- 1}$ and $Δ S = 40 J K^{- 1} m o l^{- 1} .$ Out of the values given below choose the minimum temperature above which the process will be spontaneous:

Options:

Answer:

Solution:

$∆ G = ∆ H - T ∆ S$ ....... (Gibbs Helmholtz Equation)

For a reaction to be spontaneous, $∆ G < 0$

So, $∆ H - T ∆ S < 0$

$∆ H < T ∆ S$

$\frac{∆ H}{∆ S} < T$

i.e. Minimum temp. should be $\Rightarrow \frac{∆ H}{∆ S} = \frac{200}{40} = 5 K$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Free Energy and Equilibrium

⏱ 2mℹ️ Source: PYQ_2019

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