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Chapter 6: Problem 52

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

Short Answer

Expert verified
The enthalpy change for burning 1.00 g of methane is -55.6 kJ, and the enthalpy change for burning \(1.00 \times 10^{3}\) L methane gas at 740 torr and 25ºC is -35,500 kJ.

Step by step solution

01

Calculate moles of methane burned in part a

Given mass of methane is 1.00 g. The molecular weight of methane (CH4) is 12.01 g/mol (C) + 4 * 1.01 g/mol (H) = 16.05 g/mol. We can calculate the moles of methane using the formula: Moles = (Mass)/(Molecular weight) Moles (CH4) = \( \frac{1.00}{16.05} \) = 0.0623 moles
02

Calculate enthalpy change for part a

The stoichiometry of the reaction is that 1 mole of CH4 produces 1 mole of CO2, and -891 kJ of heat is released. So for every mole of CH4 reacted, the enthalpy change is -891 kJ. We can use the ratio to determine the enthalpy change for 0.0623 moles of CH4 as follows: ΔH = -(891 kJ/mol) * (0.0623 moles) = -55.6 kJ Hence, the enthalpy change for burning 1.00 g of methane is -55.6 kJ. #b. \(1.00 \times 10^{3}\) L methane gas at 740 torr and \(25^{\circ} \mathrm{C}\) are burned in excess oxygen#
03

Calculate moles of methane burned in part b

In this case, we are given the volume, pressure, and temperature of the methane gas. We can use the ideal gas law equation (PV = nRT) to find the moles of CH4. Before we do that, we need to convert the given units to SI units: Pressure (P) = 740 torr * (101.325 kPa / 760 torr) = 98.5 kPa Temperature (T) = 25ºC + 273.15 = 298.15 K Volume (V) = 1.00 * 10^3 L Now, we can calculate the number of moles (n) using the ideal gas law equation: n = PV / RT R = 8.314 J/(mol K) n = \( \frac{98.5 \times 1000}{8.314 \times 298.15} \) = 39.9 moles
04

Calculate enthalpy change for part b

Similar to part a, we use the stoichiometry of the reaction and the enthalpy change to calculate the enthalpy change for burning 39.9 moles of CH4: ΔH = -(891 kJ/mol) * (39.9 moles) = -35,500 kJ Hence, the enthalpy change for burning \(1.00 \times 10^{3}\) L methane gas at 740 torr and 25ºC is -35,500 kJ.

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