Question

calculate the standard entropy change for the combustion of propane, a common fuel.\\(\ce{c3h8(g) + 5o2(g) -> 3co2(g) + 4h2o(l)}\\)\
\

substance	\\(s^\circ\\) (j/mol·k)	\
---	---	\
\\(\ce{c3h8(g)}\\)	270	\
\\(\ce{o2(g)}\\)	205	\
\\(\ce{co2(g)}\\)	214	\
\\(\ce{h2o(l)}\\)	70	\

\\(\delta s^\circ = ?\\) j/k\
enter either a + or - sign and the magnitude in your answer.

Explanation:

Step1: Recall the formula for standard entropy change

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

Step2: Calculate the entropy of products

For the products: $3$ moles of $\text{CO}_2(\text{g})$ and $4$ moles of $\text{H}_2\text{O}(\text{l})$.
Entropy of products: $3\times S^\circ(\text{CO}_2) + 4\times S^\circ(\text{H}_2\text{O})$
Substitute the values: $3\times214 + 4\times70 = 642 + 280 = 922$ J/K.

Step3: Calculate the entropy of reactants

For the reactants: $1$ mole of $\text{C}_3\text{H}_8(\text{g})$ and $5$ moles of $\text{O}_2(\text{g})$.
Entropy of reactants: $1\times S^\circ(\text{C}_3\text{H}_8) + 5\times S^\circ(\text{O}_2)$
Substitute the values: $1\times270 + 5\times205 = 270 + 1025 = 1295$ J/K.

Step4: Calculate $\Delta S^\circ$

$\Delta S^\circ =$ Entropy of products - Entropy of reactants $= 922 - 1295 = -373$ J/K.

Answer:

-373