Question

calculate the change in entropy for the reaction.
sicl₄(l) + 2h₂o(l) → sio₂(s) + 4hcl(aq)
substance | s° (j/mol·k)
sicl₄(l) | 240.
h₂o(l) | 70.
sio₂(s) | 42
hcl(aq) | 46
δs° = ? j/k
enter either a + or - sign and the magnitude in your answer.

Explanation:

Step1: Recall the formula for entropy change

The formula for the standard entropy change (\(\Delta S^\circ\)) of a reaction is \(\Delta S^\circ=\sum nS^\circ(\text{products})-\sum mS^\circ(\text{reactants})\), where \(n\) and \(m\) are the stoichiometric coefficients of the products and reactants respectively.

Step2: Calculate the entropy of products

For the products: \(\text{SiO}_2(\text{s})\) has a stoichiometric coefficient of 1 and \(S^\circ = 42\space\text{J/mol·K}\), and \(\text{HCl}(\text{aq})\) has a stoichiometric coefficient of 4 and \(S^\circ=46\space\text{J/mol·K}\).
So, \(\sum nS^\circ(\text{products})=(1\times42)+(4\times46)=42 + 184=226\space\text{J/K}\)

Step3: Calculate the entropy of reactants

For the reactants: \(\text{SiCl}_4(\text{l})\) has a stoichiometric coefficient of 1 and \(S^\circ = 240\space\text{J/mol·K}\), and \(\text{H}_2\text{O}(\text{l})\) has a stoichiometric coefficient of 2 and \(S^\circ = 70\space\text{J/mol·K}\).
So, \(\sum mS^\circ(\text{reactants})=(1\times240)+(2\times70)=240+140 = 380\space\text{J/K}\)

Step4: Calculate \(\Delta S^\circ\)

Using the formula \(\Delta S^\circ=\sum nS^\circ(\text{products})-\sum mS^\circ(\text{reactants})\)
\(\Delta S^\circ=226 - 380=- 154\space\text{J/K}\)

Answer:

\(-154\)