Standard entropies of X2- Y2 and XY3 are 60- 40 and 50 JK-1mol-1 respectively- For the reaction- 12 X2 - 32Y2 - XY3- H - -30 kJ-molto be at spontaneous- the temperature should be-

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Gibbs Free Energy & Spontaneity

Standard entr...

Question

Standard entropies of $X_{2}, Y_{2} a n d X Y_{3}$ are 60, 40 and $50 J K^{- 1} m o l^{- 1}$ respectively. For the reaction:
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}; △ H = - 30 k J / m o l$
to be at spontaneous, the temperature should be:

750 K

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1000 K

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1250 K

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500 K

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Solution

The correct option is A 750 K
the reaction:
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}; △ H = - 30 k J / m o l$
$△ S_{X_{2}} = 60 J K^{- 1} m o l^{- 1}$
$△ S_{Y_{2}} = 40 J K^{- 1} m o l^{- 1}$
$△ S_{X Y_{3}} = 50 J K^{- 1} m o l^{- 1}$
$△ S = \sum S_{P r o d u c t s} - \sum S_{R e a c t a n t s}$
$△ S = △ S_{X Y_{3}} - [△ S_{X_{2}} + △ S_{Y_{2}}]$
$= 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40]$
$= 50 - (30 + 60) = - 40 J K^{- 1} m o l^{- 1}$

for reaction to be spontaneous putting $△ G < 0$
so if, $△ G = △ H - T △ S < 0$
i.e., $△ H < T △ S$
As both have negative sign so inequality will change.

$T < \frac{△ H}{△ S} T < \frac{- 30 \times 1000}{- 40} T < 750 K$
below this temperature the reaction will spontaneous, because $△ G < 0$

Theory:

For spontaneous process, $Δ S_{u n i v}$ is greater than 0 and minimum value of T is 0, T will be positive.

So for spontaneity $Δ G_{s y s}$ must be negative.

For Spontaneous reaction $G_{s y s}$ is less than 0
For non spontaneous reaction $G_{s y s}$ is greater than 0
For equilibrium condition $G_{s y s} = 0$

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Gibbs Free Energy & Spontaneity

Standard XII Chemistry

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Standard entropies of X2,Y2 and XY3 are 60, 40 and 50 JK−1mol−1 respectively. For the reaction: 12X2+32Y2⇌XY3; △H=−30 kJ/mol to be at spontaneous, the temperature should be:

Standard entropies of $X_{2}, Y_{2} a n d X Y_{3}$ are 60, 40 and $50 J K^{- 1} m o l^{- 1}$ respectively. For the reaction:
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}; △ H = - 30 k J / m o l$
to be at spontaneous, the temperature should be: