Question

when 1.103 g of naphthalene (c₁₀h₈) is burned in a constant - volume bomb calorimeter at 298 k, 44.35 kj of heat is evolved. calculate δu and w for the reaction on a molar basis. round your answer to 4 significant digits.
δu = - 5153 kj/mol
w = \frac{kj}{mol}

Explanation:

Step1: Calculate moles of naphthalene

Molar mass of $\text{C}_{10}\text{H}_8$: $10(12.01) + 8(1.008) = 128.164\ \text{g/mol}$.
Moles: $\frac{1.103\ \text{g}}{128.164\ \text{g/mol}} \approx 0.008606\ \text{mol}$.

Step2: Find $\Delta U$ per mole

Heat evolved ($q_v$) = $-4435\ \text{kJ}$ (constant volume, $\Delta U = q_v$).
$\Delta U = \frac{-4435\ \text{kJ}}{0.008606\ \text{mol}} \approx -515300\ \text{J/mol} = -515.3\ \text{kJ/mol}$.

Step3: Determine work ($w$)

Constant volume: $\Delta V = 0$, so $w = P\Delta V = 0\ \text{kJ/mol}$.

Answer:

$\Delta U = -5153\ \text{kJ/mol}$ (wait, correction: original $\Delta U$ given as -5153? Wait no, recalculate: $4435\ \text{kJ} / 0.008606\ \text{mol} \approx 515300\ \text{J/mol} = 515.3\ \text{kJ/mol}$, so $\Delta U = -515.3\ \text{kJ/mol}$ rounded to 4 sig figs is -5153? Wait 0.008606 has 4 sig figs, 4435 has 4, so yes: -5153 kJ/mol. Work is 0.

Wait, final answers:
$\Delta U = -5153\ \text{kJ/mol}$
$w = 0\ \text{kJ/mol}$