Standard entropies of X2,Y2andXY3 are 60, 40 and 50JK−1mol−1 respectively. For the reaction: 12X2+32Y2⇌XY3;△H=−30kJ/mol to be at spontaneous, the temperature should be:
A
750 K
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B
1000 K
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C
1250 K
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D
500 K
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Solution
The correct option is A 750 K the reaction: 12X2+32Y2⇌XY3;△H=−30kJ/mol △SX2=60JK−1mol−1 △SY2=40JK−1mol−1 △SXY3=50JK−1mol−1 △S=∑SProducts−∑SReactants △S=△SXY3−[△SX2+△SY2] =50−[12×60+32×40] =50−(30+60)=−40JK−1mol−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<△H△ST<−30×1000−40T<750K below this temperature the reaction will spontaneous, because △G<0
Theory:
For spontaneous process, ΔSuniv is greater than 0 and minimum value of T is 0, T will be positive.
So for spontaneity ΔGsys must be negative.
For Spontaneous reaction Gsys is less than 0 For non spontaneous reaction Gsys is greater than 0 For equilibrium condition Gsys=0