
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

The standard enthalpies of formation for $\mathrm{Na}_{2} \mathrm{O}$(solid), $\mathrm{H}_{2} \mathrm{O}$(liquid), and $\mathrm{NaOH}$(solid) at 298 K are $-416$, $-286$, and $-427.8$ kJ/mol, respectively. Using Hess’s Law, calculate $\Delta H^{\circ}{ }_{298}$ for the chemical reaction:

$$\Delta H^{\mathrm{o}}{ }_{298}=2^{\Delta H_{f}^{0}}(\mathrm{NaOH}, \mathrm{s})-\left[{ }^{\Delta H_{f}^{0}}\left(\mathrm{Na}_{2} \mathrm{O}, \mathrm{s}\right)+\Delta H_{f}^{0}\left(\mathrm{H}_{2} \mathrm{O}, \mathrm{l}\right)\right]=2 \cdot(-427.8)-[-416+(-286)]=-153.6\: \text{kJ}.$$

In the thermochemical equation, the coefficients represent moles, allowing for fractional values. One mole of sodium hydroxide can be formed from $1 / 2$ mole of sodium oxide and $1 / 2$ mole of water. It was calculated that the formation of 2 moles of $\mathrm{NaOH}$ releases 153.6 kJ of heat ($\Delta H^{\mathrm{o}}{ }_{298}=-153.6$ kJ). Therefore, the formation of 1 mole $\mathrm{NaOH}$ will release half the heat, i.e., 76.8 kJ. The thermochemical equation for this formation, specifying the released heat with a positive sign, is:

$$\frac{1}{2} \mathrm{Na}_{2} \mathrm{O}(\mathrm{s}) + \frac{1}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NaOH}(\mathrm{s}) + 76.8 \: \text{kJ}.$$

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