Question

H2S(g) 2H2O(l)3H2(g) SO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.60 moles of H2S(g) react at standard conditions.

Answer 1

Answer:
`\Delta S` = 473.92J/K.mol

Step-by-step explanation: In physics, Entropy is defined as a degree of disorder in a system. Entropy change is given by the sum of all the products multiplied by their respective coeficients minus the sum of all the reagents multiplied by their respective coeficients:

`\Delta S = m\Sigma product - n\Sigma reagent`

The balanced reaction:

`H_(2)S_((g))+2H_(2)O_((l))=>3H_(2)_((g))+SO_(2)_((g))`

gives the proportion reagents react to form products, so, if 1.6 moles of
`H_(2)S_((g))`:

3.2 moles of water is used;

4.8 moles of hydrogen gas is formed;

1.6 moles of sulfur dioxide is also formed;

Calculating entropy change:

`\Delta S = (4.8*131+1.6*248.8)-(1.6*205.6+3.2*70)`

`\Delta S=628.8+398.08-328.96-224`

`\Delta S` = 473.92J/K.mol

Entropy change for the given chemical reaction is
`\Delta S` = 473.92J/K.mol