Question

For the given reaction
$H_2\left(g\right)+S\left(s\right)⟶H_{2}S\left(g\right);\Delta H_T=100kJ/mol$ and $\Delta S_T=400J/mol/K$
Temperature at which above reaction occurs reversibly is (Assuming $\Delta H_T$ and $\Delta S_T$ are independent of temperature)

• A. $200K$
• B. $250K$
• C. $400K$
• D. $None$

Answer 1

Answer: (B)

$250K$

We know that

$\Delta S_T=\frac{\Delta H_T}{T}$

Given, $\Delta S_T=400Jmo{l}^{-1}{K}^{-1}$

$\Delta H_T=100KJmo{l}^{-1}$

So $T=\frac{\Delta H_T}{\Delta S_T}=\frac{100\times 1000}{400}=250K$

Hence, Temperature is $250K$ .