Answer:
[tex]\mathbf{\Delta \ S^0 \ system \simeq -71.862 \ J/K}[/tex]
Explanation:
The chemical equation for the reaction is given as:
[tex]\mathsf{4 H Cl _{(g)} + O_{2(g)} \to 2H_2O _{(g)} + 2Cl_{2(g)}}[/tex]
the entropy change in the system can be calculated as follows:
[tex]\Delta S^0 \ system = \Delta S(products) - \Delta S (reactants)[/tex]
[tex]\Delta S^0 \ system = (2 \times \Delta S(H_2O )+2 \times \Delta S(Cl_2 ) ) - (4 \times \Delta S (HCl) + 1 \times \Delta S (O_2))[/tex]
From the tables; the entropy values where obtained.
∴
[tex]\Delta S^0 \ system = (2 \times(188.8 \ J/K )+2 \times ( 223.1 \ J/K )) - (4 \times \ ( 186.9 \ J/K ) + 1 \times ( 205.1 \ J/K ))[/tex]
[tex]\Delta S^0 \ system = (377.6 \ J/K +446.2 \ J/K )) - (747.6\ J/K ) + 205.1 \ J/K ))[/tex]
[tex]\Delta S^0 \ system = (377.6 \ J/K +446.2 \ J/K - 747.6\ J/K - 205.1 \ J/K )[/tex]
[tex]\Delta S^0 \ system = (-128.9 \ J/K )[/tex]
i.e the entropy change in the system when 4 moles of HCl is used = -128.9 J/K
∴
when 2.23 moles of HCl is used, Then,
[tex]\Delta \ S^0 \ system = \dfrac{-128.9 \ J/K }{4 \ mol } \times 2.23 \ mol[/tex]
[tex]\Delta \ S^0 \ system = -32.225 \ J/K \times 2.23[/tex]
[tex]\Delta \ S^0 \ system = -71.86175 \ J/K[/tex]
[tex]\mathbf{\Delta \ S^0 \ system \simeq -71.862 \ J/K}[/tex]
