A pellet of naphthalene of mass 1.28 g is burnt in a bomb calorimeter with heat capacity 14000 J/K. If the initial temperature is 300K and final temperature is 302K.
The $| Δ H_{r e a c t i o n} |$ for 1 mole of napthalene at 300 K in KJ/mole (consider reactant & products at 300K).
[At. wt. of C = 12, $C_{V, C O_{2} (g)} = 35 J m o l^{- 1} K^{- 1}$ and $R = 8.3 J m o l^{- 1} K^{- 1}$
At. wt. of H = 1, $C_{V, H_{2} O (i)} = 75 J m o l^{- 1} K^{- 1},$ heat capacity of $C O_{2} (g)$ & $H_{2} O (l)$ are taken as constant in this temperature range].

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Solution

Given, Heat capacity of calorimeter =14000 J/K
$T_{i} = 300 K, T_{f} = 302 K$
$C_{V, C O_{2} (g)} = 35 J m o l^{- 1} K^{- 1}$ and $R = 8.3 J m o l^{- 1} K^{- 1}$
$C_{V, H_{2} O (i)} = 75 J m o l^{- 1} K^{- 1}$
$C_{10} H_{8} (s) + 12 O_{2} (g) \to 10 C O_{2} (g) + 4 H_{2} O (l)$
For 1 mole Napthalene, heat liberated = $\frac{14000 \times 2}{0.01} = 28 \times 10^{5} J / m o l e$
$Δ \cup = - 28 \times 10^{5} J / m o l$
$Δ \cup_{at 300 K} = - 28 \times 10^{5} - [10 \times 2 \times 35 + 4 \times 75 \times 2]$
(but here product and reactant both are at 300K)
$= (- 28 \times 10^{5} - 1300) J$
$= (- 2800 \times 1.3) k J = - 2801.3 k J$
$Δ H = Δ U + (Δ n)_{g} R T$
$Δ H = - 2801.3 + (- 2) 8.3 \times 300 \times 10^{- 3}$
$Δ H = 2806.28 K J / m o l$

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