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For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), Δ H = -890.3 kJ mol⁻¹ at 298 K…

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For $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ , $\Delta H = -890.3\ \text{kJ mol}^{-1}$ at $298\ \text{K}$ . Find $\Delta n_g$ and then $\Delta U$ . Would a bomb calorimeter give out more heat or less than an open flask?

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…

The worked solution

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