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

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

Step-by-step 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 * \Delta S(H_2O )+2 * \Delta S(Cl_2 ) ) - (4 * \Delta S (HCl) + 1 * \Delta S (O_2))`

From the tables; the entropy values where obtained.

∴

`\Delta S^0 \ system = (2 *(188.8 \ J/K )+2 * ( 223.1 \ J/K )) - (4 * \ ( 186.9 \ J/K ) + 1 * ( 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 = (-128.9 \ J/K )/(4 \ mol ) * 2.23 \ mol`

`\Delta \ S^0 \ system = -32.225 \ J/K * 2.23`

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

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