For a given reaction- - H -35-5 kJ mol -1 and - S -83-6 J K -1 mol -1 - The reaction is spontaneous at Assume that - H and - S do not vary with temperature- A- T -425 KB- All temperatures C- T 298 K

For a spontaneous reaction, Δ G lt; 0 i.e., Δ H T Δ S lt; 0 T gt; Δ H/Δ S T gt; ( 35.5 × 1000/83.6=424.6 ≈ 425 K ) ∴ T gt; 425K ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

For a given r...

Question

For a given reaction, $Δ H = 35.5 k J m o l^{- 1} a n d Δ S = 83.6 J K^{- 1} m o l^{- 1}$ . The reaction is spontaneous at (Assume that $Δ H a n d Δ S$ do not vary with temperature.)

T > 425 K

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All temperatures

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T > 298 K

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T < 425 K

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Solution

The correct option is A T > 425 K
For a spontaneous reaction,
$Δ G < 0 i . e ., Δ H - T Δ S < 0$
$T > \frac{Δ H}{Δ S}$
$T > (\frac{35.5 \times 1000}{83.6} = 424.6 \approx 425 K)$
$∴ T > 425 K$

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For a given reaction, ΔH=35.5 kJ mol−1 and ΔS=83.6J K−1mol−1. The reaction is spontaneous at (Assume that ΔH and ΔS do not vary with temperature.)

The correct option is A T > 425 KFor a spontaneous reaction, ΔG<0 i.e., ΔH−TΔS<0 T>ΔHΔS T>(35.5×100083.6=424.6≈425K) ∴ T>425K

For a given reaction, $Δ H = 35.5 k J m o l^{- 1} a n d Δ S = 83.6 J K^{- 1} m o l^{- 1}$ . The reaction is spontaneous at (Assume that $Δ H a n d Δ S$ do not vary with temperature.)

The correct option is A T > 425 K
For a spontaneous reaction,
$Δ G < 0 i . e ., Δ H - T Δ S < 0$
$T > \frac{Δ H}{Δ S}$
$T > (\frac{35.5 \times 1000}{83.6} = 424.6 \approx 425 K)$
$∴ T > 425 K$