wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Standard entropies of X2,Y2 and XY3 are 60, 40 and 50 JK1mol1 respectively. For the reaction:
12X2+32Y2XY3; H=30 kJ/mol
to be at spontaneous, the temperature should be:

A
750 K
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1000 K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1250 K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
500 K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 750 K
the reaction:
12X2+32Y2XY3; H=30 kJ/mol
SX2=60 JK1mol1
SY2=40 JK1mol1
SXY3=50 JK1mol1
S=SProductsSReactants
S=SXY3[SX2+SY2]
=50[12×60+32×40]
=50(30+60)=40 JK1mol1

for reaction to be spontaneous putting G<0
so if, G=HTS<0
i.e., H<TS
As both have negative sign so inequality will change.

T<HST<30×100040T<750 K
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

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon