QUESTION IMAGE
Question
using the data below, calculate the entropy for the formation of rust, iron(ii) oxide from fe and h₂o. 2fe(s) + 3h₂o(g) → fe₂o₃(s) + 3h₂(g) \
$\delta s_\text{rxn} = ?$ j/k enter either a + or - sign and the magnitude.
Step1: Recall the formula for entropy change of reaction
The formula for the entropy change of a reaction ($\Delta S_{rxn}$) is the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants, each multiplied by their stoichiometric coefficients. Mathematically, it is:
$$\Delta S_{rxn}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})$$
where $n$ and $m$ are the stoichiometric coefficients of products and reactants respectively.
Step2: Identify the stoichiometric coefficients and standard entropies
For the reaction $2\text{Fe}(s) + 3\text{H}_2\text{O}(g)
ightarrow\text{Fe}_2\text{O}_3(s)+3\text{H}_2(g)$:
- Reactants: $\text{Fe}(s)$ with $n = 2$ and $S^{\circ}=15\ \text{J/mol K}$; $\text{H}_2\text{O}(g)$ with $n = 3$ and $S^{\circ}=189\ \text{J/mol K}$.
- Products: $\text{Fe}_2\text{O}_3(s)$ with $n = 1$ and $S^{\circ}=87\ \text{J/mol K}$; $\text{H}_2(g)$ with $n = 3$ and $S^{\circ}=131\ \text{J/mol K}$.
Step3: Calculate the sum of entropies of products
Sum of entropies of products ($\sum nS^{\circ}(\text{products})$):
$$S_{\text{products}}=(1\times87)+(3\times131)$$
$$S_{\text{products}} = 87+393$$
$$S_{\text{products}}=480\ \text{J/K}$$
Step4: Calculate the sum of entropies of reactants
Sum of entropies of reactants ($\sum mS^{\circ}(\text{reactants})$):
$$S_{\text{reactants}}=(2\times15)+(3\times189)$$
$$S_{\text{reactants}}=30 + 567$$
$$S_{\text{reactants}}=597\ \text{J/K}$$
Step5: Calculate $\Delta S_{rxn}$
Using the formula $\Delta S_{rxn}=S_{\text{products}}-S_{\text{reactants}}$:
$$\Delta S_{rxn}=480 - 597$$
$$\Delta S_{rxn}=- 117\ \text{J/K}$$
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