Consider the following reaction: CH4 +2O2 -> CO2 + 2H2O. Delta H = -891 kJ
Calculate the enthalpy change for each of the following cases:
a. 1.00 g methane is burned in excess oxygen
b. 1.00 x 10^3 L methane gas at 740. torr and 25 *C is burned in excess oxygen
To calculate the enthalpy change for each case, we need to use the concept of stoichiometry and the given enthalpy change for the reaction.
a. To calculate the enthalpy change when 1.00 g of methane is burned in excess oxygen, we need to determine the number of moles of methane being burned. Here's how to do it:
1. Calculate the molar mass of methane (CH4). The atomic masses are: C = 12.01 g/mol, H = 1.01 g/mol.
Molar mass of CH4 = (12.01 g/mol x 1) + (1.01 g/mol x 4) = 16.05 g/mol
2. Convert the given mass of methane into moles using the molar mass:
Moles of CH4 = (1.00 g / 16.05 g/mol) = 0.0623 mol
3. Since the reaction shows that 1 mole of methane reacts with 2 moles of oxygen, we need to calculate the number of moles of oxygen used in the reaction:
Moles of O2 = 2 x Moles of CH4 = 2 x 0.0623 mol = 0.1246 mol
4. Now we can calculate the enthalpy change (ΔH) using the given value:
Enthalpy change = ΔH x Moles of CH4 = -891 kJ/mol x 0.0623 mol = -55.5 kJ
Therefore, the enthalpy change for case a is -55.5 kJ.
b. To calculate the enthalpy change when 1.00 x 10^3 L of methane gas at 740. torr and 25 °C is burned in excess oxygen, we need to consider the ideal gas law and the stoichiometry of the reaction.
1. Convert the given volume of methane gas at 25 °C and 740. torr into moles using the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Convert the pressure to atm: 740. torr = 740. torr / 760 torr/atm = 0.974 atm
Convert the temperature to Kelvin: 25 °C = 25 + 273.15 = 298.15 K
Using the ideal gas law: (0.974 atm) x V = n x (0.0821 L·atm/(mol·K)) x (298.15 K)
n = (0.974 atm x V) / (0.0821 L·atm/(mol·K) x 298.15 K)
2. Calculate the number of moles of methane gas using the given volume:
Moles of CH4 = (0.974 atm x 1.00 x 10^3 L) / (0.0821 L·atm/(mol·K) x 298.15 K)
3. Calculate the number of moles of oxygen used in the reaction:
Moles of O2 = 2 x Moles of CH4
4. Calculate the enthalpy change (ΔH) using the given value:
Enthalpy change = ΔH x Moles of CH4
Therefore, to calculate the enthalpy change for case b, you'll need to use the ideal gas law and stoichiometry to determine the number of moles of methane gas, and then proceed with the calculation described above.
To calculate the enthalpy change for each case, we will use the concept of stoichiometry and the given molar enthalpy change.
a. 1.00 g methane is burned in excess oxygen:
To begin, we need to convert the mass of methane (CH4) to moles. The molar mass of methane (CH4) is 16.04 g/mol (1 x 12.01 g/mol for carbon + 4 x 1.01 g/mol for hydrogen).
Mass of CH4 = 1.00 g
Molar mass of CH4 = 16.04 g/mol
Number of moles of CH4 = Mass of CH4 / Molar mass of CH4
Number of moles of CH4 = 1.00 g / 16.04 g/mol ≈ 0.0623 mol
From the balanced equation:
1 mol CH4 produces -891 kJ of energy
Therefore, the enthalpy change for 0.0623 mol CH4 is:
Enthalpy change (ΔH) = Number of moles of CH4 x ΔH per mole
Enthalpy change (ΔH) = 0.0623 mol x -891 kJ/mol = -55.5 kJ
So, the enthalpy change for 1.00 g of methane burned in excess oxygen is approximately -55.5 kJ.
b. 1.00 x 10^3 L methane gas at 740. torr and 25 °C is burned in excess oxygen:
To calculate the enthalpy change for this case, we need to first convert the volume of methane gas (CH4) to the number of moles. To do this, we'll use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
We need to convert the given pressure (740. torr) and temperature (25 °C) to atm and Kelvin, respectively.
Pressure (P) = 740. torr / 760 torr/atm ≈ 0.974 atm
Temperature (T) = 25 °C + 273.15 = 298.15 K
Using the ideal gas law equation, we can now solve for the number of moles (n) of CH4:
n = PV / RT
n = (0.974 atm) x (1.00 x 10^3 L) / (0.0821 L.atm/mol.K) x (298.15 K)
n ≈ 40.78 mol
From the balanced equation:
1 mol CH4 produces -891 kJ of energy
Therefore, the enthalpy change for 40.78 mol CH4 is:
Enthalpy change (ΔH) = Number of moles of CH4 x ΔH per mole
Enthalpy change (ΔH) = 40.78 mol x -891 kJ/mol = -36,332 kJ
So, the enthalpy change for burning 1.00 x 10^3 L of methane gas at 740. torr and 25 °C in excess oxygen is approximately -36,332 kJ.
a.
-891 kJ x 1/16 = ??
b.
Use PV = nRT and solve for n, convert to grams, and use the same system as in a; i.e.,
-891 kJ x (grams/16) = ??