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 of a reaction

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: Identify the products and their stoichiometric coefficients

The products are $\text{SiO}_2(s)$ with a stoichiometric coefficient of 1 and $\text{HCl}(aq)$ with a stoichiometric coefficient of 4.
So, $\sum nS^\circ(\text{products}) = 1\times S^\circ(\text{SiO}_2(s))+4\times S^\circ(\text{HCl}(aq))$
Substitute the values: $1\times42 + 4\times46=42 + 184 = 226$ $\text{J/K}$

Step3: Identify the reactants and their stoichiometric coefficients

The reactants are $\text{SiCl}_4(l)$ with a stoichiometric coefficient of 1 and $\text{H}_2\text{O}(l)$ with a stoichiometric coefficient of 2.
So, $\sum mS^\circ(\text{reactants})=1\times S^\circ(\text{SiCl}_4(l)) + 2\times S^\circ(\text{H}_2\text{O}(l))$
Substitute the values: $1\times240+2\times70 = 240 + 140=380$ $\text{J/K}$

Step4: Calculate the entropy change

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

Answer:

-154