Question

A chemist claims that the following reaction is feasible at 298 K
SF(8) + 8 HI (8) -> H2S (8 + 6HF (8) + 4 12 (s)]
Verify his claim. Given that AGfor SF6 (8), HI (8), H2S (8) and HF (8) are - 991.61, 1.30, – 33-01 and
- 270.73 kJ mol-1 respectively.
​

Answer 1

Answer: The Gibbs free energy of the reaction is coming out to be negative.

Explanation:

For the given chemical reaction:

$SF_6(g)+8HI(g)\rightarrow H_2S(g)+6HF(g)+4I_2(s)$

The equation used to calculate Gibbs free change of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta G^o_{rxn}=[(1\times \Delta G^o_f_{(H_2S(g))})+(6\times \Delta G^o_f_{(HF(g))})+(4\times \Delta G^o_f_{(I_2(s))})]-[(1\times \Delta G^o_f_{(SF_6(g))})+(8\times \Delta G^o_f_{(HI(g))})]$

We are given:

$\Delta G^o_f_{(H_2S(g))}=-33.01kJ/mol\\\Delta G^o_f_{(HF(g))}=-270.73kJ/mol\\\Delta G^o_f_{(I_2(s))}=0kJ/mol\\\Delta G^o_f_{(SF_6(g))}=-991.61kJ/mol\\\Delta G^o_f_{(HI(g))}=1.30kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(1\times (-33.01))+(6\times (-270.73))+(4\times (0))]-[(1\times (-991.61))+(8\times (1.30))]=kJ/mol$

$[(1\times (-33.01))+(6\times (-270.73))+(4\times (0))]-[(1\times (-991.61))+(8\times (1.30))]=-676.18kJ$

For the reaction to be feasible, the Gibbs free energy of the reaction must come out to be negative.

From the above calculation, the Gibbs free energy of the reaction is negative and thus the reaction is feasible in nature.