Consider the following reaction.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) ÄH = -891 kJ
Calculate the enthalpy change for each of the following cases.
1.00 103 L methane gas at 719 torr and 26°C is burned in excess oxygen.
I would use PV = nRT and solve for n = moles CH4. Then
891 kJ/mol x n moles = heat released.
To calculate the enthalpy change for the given reaction, we need to use the stoichiometry of the reaction and the given conditions.
1. First, we need to convert the given volume of methane gas from liters to moles. We can use the ideal gas law equation to do this.
PV = nRT
Where:
P = pressure in torr
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
Given:
P = 719 torr
V = 1.00 × 10^3 L
T = 26 °C = 299 K
Plugging in the values into the ideal gas law equation:
(719 torr) × (1.00 × 10^3 L) = n × (0.0821 L·atm/(mol·K)) × (299 K)
n = (719 × 1.00 × 10^3)/(0.0821 × 299) moles
n ≈ 30.72 moles
2. Since methane gas is burned in excess oxygen, it means that all of the methane gas will react. Therefore, the stoichiometry of the reaction tells us that for every 1 mole of methane burned, we get 891 kJ of energy.
So, the enthalpy change for burning 30.72 moles of methane gas will be:
Enthalpy Change = 30.72 moles × (-891 kJ/mol)
Enthalpy Change ≈ -27338.72 kJ
Therefore, the enthalpy change for burning 1.00 × 10^3 L of methane gas at 719 torr and 26°C in excess oxygen is approximately -27338.72 kJ.