The heat of combination of napthalene, $C_{10} H_{8} (s)$ , at constant volume at $25^{o} C$ was found to be $- 5133 k J m o l^{- 1}$ . Calculate type value of enthalpy change at constant pressure at the same temperature.

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Solution

${C_{10} H_{8}}_{(s)} + 12 {O_{2}}_{(g)} ⟶ 10 {C O_{2}}_{(g)} + 4 {H_{2} O}_{(l)}$
For the above reaction-
$Δ n_{g} = n_{P} - n_{R} = 10 - 12 = - 2$
$Δ E = - 5133 k J / m o l (Given)$
$T = 25 ℃ = (25 + 273) K = 298 K$
Now, as we know that
$Δ H = Δ E + Δ n_{g} R T$
$\Rightarrow Δ H = - 5133 + (- 2 \times 8.314 \times 10^{- 3} \times 298)$
$\Rightarrow Δ H = - 5133 - 4.95 = - 5137.95 k J / m o l$
Hence the enthalpy change at constant pressure is $- 5137.95 k J / m o l$ .

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