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Question

$1 g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298 K$ and $1$ atmospheric pressure according to the equation

$C_{graphite} + O_{2} (g) ⟶ C O_{2} (g)$

During the reaction, temperature rises from $298 K$ 𝑡𝑜 $299 K .$ If the heat capacity of the bomb calorimeter is $20.7 k J / K,$ what is the enthalpy change for the above reaction at $298 K$ and $1 a t m ?$

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Solution

Step 1: Released Heat

let $q$ be the quantity of heat and $C_{V}$ be the heat capacity of the calorimeter.

Then, the heat absorbed by the calorimeter:

$q = C_{V} \times △ T$

Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat released by the system (reaction mixture) is equal to the heat absorbed by the calorimeter.

$q = - C_{v} \times △ T = - 20.7 k J / K \times (299 - 298) K = - 20.7 k J$

(Here, negative sign indicates the exothermic nature of the reaction)

Thus, $△ U$ for the combustion of 1g of graphite $= - 20.7 k J K^{- 1}$

$△ U$ for combustion of $1 m o l$ of graphite $= \frac{12 g m o l^{- 1} \times (- 20.7 k J K^{- 1})}{1 g}$

$△ U = - 2.48 \times 10^{2} k J m o l^{- 1}$

Step 2: Enthalpy change

We know, enthalpy change $(△ H) = △ U + △ n_{g} R T$

Given, $△ T = 298 K$

From the given reaction, $△ n_{g} = (1 - 1 + 0) = 0$

$△ H = △ U = - 2.48 \times 10^{2} k J m o l^{- 1}$

Final answer: $△ H = - 2.48 \times 10^{2} k J m o l^{- 1}$

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