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0.600 g of graphite burns completely in a bomb calorimeter whose heat capacity is 12.5 kJ K⁻¹…

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$0.600\ \text{g}$ of graphite burns completely in a bomb calorimeter whose heat capacity is $12.5\ \text{kJ K}^{-1}$ . The temperature rises by $1.58\ \text{K}$ . Find $\Delta U$ for burning one mole of carbon, with $12.0\ \text{g mol}^{-1}$ as its molar mass. Then find $\Delta H$ for $\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$ .

Need a reminder? The key ideaShow

At constant pressure the heat taken in is $\Delta H$ , and at constant volume it is $\Delta U$ . For gases, $\Delta H = \Delta U + \Delta n_g R T$ , where $\Delta n_g$ counts moles of gas only and $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ . Heat and temperature are linked by $q = C\Delta T = m c \Delta T$ . A calorimeter takes up the heat the reaction gives out, so $\Delta H = -q$ per mole. For an ideal gas, $C_p - C_v = R$ .

The worked solution

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← For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), Δ H = -890.3 kJ mol⁻¹ at 298 K…Argon is close to an ideal gas, with Cv = 12.47 J K⁻¹mol⁻¹… →

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0.600 g of graphite burns completely in a bomb calorimeter whose heat capacity is 12.5 kJ K⁻¹…

The worked solution

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