Estimate the boiling point of Br2(l) ( H = 30.9 kJ; S = 93.0 J/K). Br2(l) Br2(g)
Answers
Answer:
The temperature at which the process be spontaneous is calculated as follows
delta G = delta H -T delta S
let delta G be =0
therefore delta H- T delta s =0
therefore T= delta H/ delta S
convert 31 Kj to J = 31 x1000= 31000 j/mol
T=31000j/mol /93 j/mol.k =333.33K
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Explanation:
The boiling point of Br2(l) is 332.25K.
Given:
Br2(l) H = 30.9 kJ; S = 93.0 J/K. Br2(l)-----> Br2(g).
To Find:
The boiling point of Br2(l).
Solution:
To find the boiling point of Br2(l) we will follow the following steps:
As we know,
The formula relation between temperature and entropy is
∆G = ∆H - T∆S
Here,
∆G is Gibbs free energy and is equal to zero because boiling is an equilibrium reaction.
For equilibrium reaction ∆G = 0
So,
0 = ∆H - T∆S
∆H = T∆S
Henceforth, the boiling point of Br2(l) is 332.25K.
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