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

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