Question

minimum temperature above which the process will became spontaneous if ∆H=10KJ mol-1 while ∆s=10JK-1 is​

Answer 1

Answer:

1k

Explanation:

beacuse minimum temperature of 0°c is 1K

Answer 2

We have to find the minimum temperature above which the process will become spontaneous if ∆H = 10 kJ/mol and ∆S = 10 J/K

solution : Gibbs free energy is given by, ∆G = ∆H - T∆S

a reaction will become spontaneous only when value of Gibbs free is negative.

i.e., ∆G < 0

⇒∆G = ∆H - T∆S < 0

⇒∆H < T∆S

here ∆S > 0

⇒T > ∆H/∆S

here given, ∆H = 10 kJ/mol = 10,000 J/mol [ ∵ 1 kJ/mol = 1000 J/mol ]

∆S = 10 J/K

so, T > 10000/10

⇒T > 1000

Therefore the minimum temperature is 1000K above which the process will become spontaneous.

also read similar questions : For the reaction at 298 K 2 A+ B<===> C, ◇H=400 kJ mol and ◇S=0.25 kJ K-1 mol-1

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