Question

using the standard formation enthalpies that follow, calculate the standard enthalpy change for this reaction. 2na(s) + 2h₂o(l) → 2naoh(aq) + h₂(g) species δh° (kj/mol) h₂o(l) -285.8 naoh(aq) -470.1 3 item attempts remaining

Explanation:

Step1: Recall enthalpy - change formula

The standard enthalpy change of a reaction $\Delta_{r}H^{\circ}=\sum n\Delta_{f}H^{\circ}_{products}-\sum m\Delta_{f}H^{\circ}_{reactants}$, where $n$ and $m$ are the stoichiometric coefficients of products and reactants respectively. For the reaction $2Na(s)+2H_{2}O(l)
ightarrow2NaOH(aq) + H_{2}(g)$, the standard - formation enthalpy of $Na(s)$ and $H_{2}(g)$ is $0$ kJ/mol by definition.

Step2: Identify products and reactants

The products are $2$ moles of $NaOH(aq)$ and $1$ mole of $H_{2}(g)$, and the reactants are $2$ moles of $Na(s)$ and $2$ moles of $H_{2}O(l)$.

Step3: Calculate $\Delta_{r}H^{\circ}$

$\Delta_{r}H^{\circ}=2\times\Delta_{f}H^{\circ}(NaOH(aq)) + 1\times\Delta_{f}H^{\circ}(H_{2}(g))-2\times\Delta_{f}H^{\circ}(Na(s))-2\times\Delta_{f}H^{\circ}(H_{2}O(l))$. Substitute $\Delta_{f}H^{\circ}(NaOH(aq))=- 470.1$ kJ/mol, $\Delta_{f}H^{\circ}(H_{2}(g)) = 0$ kJ/mol, $\Delta_{f}H^{\circ}(Na(s)) = 0$ kJ/mol, and $\Delta_{f}H^{\circ}(H_{2}O(l))=-285.8$ kJ/mol into the formula.

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Answer:

$-368.6$