Question

Consider the reaction: 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.23 moles of HCl(g) react at standard conditions. S°system = J/K

Answer 1

Answer:

$\mathbf{\Delta \ S^0 \ system \simeq -71.862 \ J/K}$

Explanation:

The chemical equation for the reaction is given as:

$\mathsf{4 H Cl _{(g)} + O_{2(g)} \to 2H_2O _{(g)} + 2Cl_{2(g)}}$

the entropy change in the system can be calculated as follows:

$\Delta S^0 \ system = \Delta S(products) - \Delta S (reactants)$

$\Delta S^0 \ system = (2 \times \Delta S(H_2O )+2 \times \Delta S(Cl_2 ) ) - (4 \times \Delta S (HCl) + 1 \times \Delta S (O_2))$

From the tables; the entropy values where obtained.

∴

$\Delta S^0 \ system = (2 \times(188.8 \ J/K )+2 \times ( 223.1 \ J/K )) - (4 \times \ ( 186.9 \ J/K ) + 1 \times ( 205.1 \ J/K ))$

$\Delta S^0 \ system = (377.6 \ J/K +446.2 \ J/K )) - (747.6\ J/K ) + 205.1 \ J/K ))$

$\Delta S^0 \ system = (377.6 \ J/K +446.2 \ J/K - 747.6\ J/K - 205.1 \ J/K )$

$\Delta S^0 \ system = (-128.9 \ J/K )$

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,

$\Delta \ S^0 \ system = \dfrac{-128.9 \ J/K }{4 \ mol } \times 2.23 \ mol$

$\Delta \ S^0 \ system = -32.225 \ J/K \times 2.23$

$\Delta \ S^0 \ system = -71.86175 \ J/K$

$\mathbf{\Delta \ S^0 \ system \simeq -71.862 \ J/K}$