Final answer:
To find the heat produced when 4.46 L of SO2 reacts to form SO3, we use the ideal gas law to calculate the moles of SO2, which is found to be 0.1809 moles. This amount of moles corresponds to -17.89 kJ of energy released during the reaction, as it is exothermic.
Step-by-step explanation:
The reaction between gaseous sulfur dioxide (SO2) and oxygen (O2) to form sulfur trioxide (SO3) involves a change in enthalpy, represented by ΔH°. The information given describes the reaction as:
2SO2 (g) + O2 (g) → 2SO3 (g)
ΔH° for this reaction is -197.8 kJ. This means the reaction is exothermic and releases 197.8 kJ of energy for every two moles of SO2 that reacts.
To calculate the amount of energy released when 4.46 L of SO2 reacts at 1.00 bar and 25.0 °C, we use the ideal gas law to find the number of moles of SO2. At standard conditions (1 bar and 0 °C), 1 mole of gas occupies 22.71 L. However, since the temperature is 25.0 °C, we must adjust for this by using the ideal gas law:
PV = nRT
Where P is pressure, V is volume, n is number of moles, R is the ideal gas constant (0.08314 L·bar/K·mol), and T is temperature in Kelvin (298 K for 25.0 °C).
By substituting the values into the ideal gas law, we can solve for n (moles of SO2):
n = PV / RT = (1.00 bar × 4.46 L) / (0.08314 L·bar/K·mol × 298 K) = 0.1809 moles of SO2
Since the ΔH° given is for two moles of SO2, we have to adjust it for 0.1809 moles:
Energy released = (0.1809 moles SO2/2 moles) × -197.8 kJ = -17.89 kJ
The negative sign indicates that energy is released in the reaction. The heat produced when 4.46 L of SO2 is converted to SO3 under the given conditions is 17.89 kJ.