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Standard XII

Chemistry

Thermodynamic Processes

A process has...

Question

A process has $Δ H = 200 J m o l^{- 1}$ and $Δ S = 40 J K^{- 1} m o l^{- 1}$ . The minimum temperature at which the process will be spontaneous is

12K

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20K

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20K

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Solution

The correct option is D
5K

For a process to be spontaneous, the Gibb's free energy has to be negative.

So, $Δ G < 0$

$Δ H - T Δ S < 0$

Given $Δ H = 200 J m o l^{- 1}$ and $Δ S = 40 J K^{- 1} m o l^{- 1}$

Substituting these values, we get minimum temperature for feasibility as 5K.

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A process has ΔH=200Jmol−1 and ΔS=40JK−1mol−1. The minimum temperature at which the process will be spontaneous is

The correct option is D 5K For a process to be spontaneous, the Gibb's free energy has to be negative. So, ΔG<0 ΔH − TΔS<0 Given ΔH=200Jmol−1 and ΔS=40JK−1mol−1 Substituting these values, we get minimum temperature for feasibility as 5K.

A process has $Δ H = 200 J m o l^{- 1}$ and $Δ S = 40 J K^{- 1} m o l^{- 1}$ . The minimum temperature at which the process will be spontaneous is

The correct option is D
5K

For a process to be spontaneous, the Gibb's free energy has to be negative.

So, $Δ G < 0$

$Δ H - T Δ S < 0$

Given $Δ H = 200 J m o l^{- 1}$ and $Δ S = 40 J K^{- 1} m o l^{- 1}$

Substituting these values, we get minimum temperature for feasibility as 5K.