Question

Calculate the entropy of $B{r}_2\left(g\right)$ in the reaction $H_2\left(g\right)+B{r}_2\left(g\right)\to 3HBr\left(g\right),△S^0=20.1J{K}^{-1}$ given entropy of $H_2$ and HBr is $130.6$ and $198.5J{K}^{-1}$.

• A. $246.3J{K}^{-1}$
• B. $123.15J{K}^{-1}$
• C. $20J{K}^{-1}$
• D. $24.63J{K}^{-1}$

Answer 1

Answer: (A)

$246.3J{K}^{-1}$

Given:

Entropy of $H_2=130.6J/K$

Entropy of HBr =198.5 J/K

$\delta S°=20.1J/K$

Solution:

Balanced chemical reaction is:

$H_2+B{r}_2\to 2HBr$

$\delta S°=2\times 198.5-130.6-EntropyofB{r}_2$

$\Rightarrow 20.1=397-130.6-EntropyofB{r}_2$

$\Rightarrow EntropyofB{r}_2=397-130.6-20.1J/K$

Therefore, Entropy of Br_2 = 246.3 J/K.

Hence, the correct option is A.