Consider the following reaction: \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-891 \mathrm{~kJ}\) Calculate the enthalpy change for each of the following cases: a. \(1.00 \mathrm{~g}\) methane is burned in excess oxygen. b. \(1.00 \times 10^{3} \mathrm{~L}\) methane gas at 740 . torr and \(25^{\circ} \mathrm{C}\) are burned in excess oxygen.

To calculate the enthalpy change when 1.00 g of methane is burned in excess oxygen, first, we need to find the number of moles of methane (n) using its molar mass: 1.00 g CH4 * \(\frac{1 \text{ mol CH4}}{16.04 \text{ g CH4}}\)= 0.0624 mol CH4 Now, using the stoichiometry of the balanced chemical equation and the standard enthalpy change for the reaction, we can calculate the enthalpy change for the reaction: Enthalpy change = \(\Delta H_{rxn}\times\frac{mol\; of\; CH_{4}}{mol\;reaction}\) Enthalpy change =(-891 kJ/mol) × (0.0624 mol CH4) Enthalpy change = -55.6 kJ

First, we need to convert the given pressure from torr to atm: P(atm) = \(\frac{740 \text{ torr}}{760 \text{ torr/atm}}\) = 0.974 atm Next, we need to convert the given temperature to Kelvin: T(K) = 25°C + 273.15 = 298.15 K Now, we can use the Ideal Gas Law (PV = nRT) to find the number of moles of methane: n = \(\frac{PV}{RT}\) n = \(\frac{(0.974 \text{ atm})(1.00 \times 10^{3} \text{ L})}{(0.0821 \frac{\text{ L atm}}{\text{K mol}})(298.15 \text{ K})}\) n = 39.86 mol CH4 Now, using the stoichiometry of the balanced chemical equation and the standard enthalpy change for the reaction, we can calculate the enthalpy change for the reaction: Enthalpy change = \(\Delta H_{rxn}\times\frac{mol\; of\; CH_{4}}{mol\;reaction}\) Enthalpy change = (-891 kJ/mol) × (39.86 mol CH4) Enthalpy change = -35,500 kJ The enthalpy changes for burning 1.00 g of methane in excess oxygen and 1.00 x 10³ L methane gas at 740 torr and 25°C in excess oxygen are -55.6 kJ and -35,500 kJ, respectively.