What is the change in enthalpy when 4.00 mol of sulfer trixoide decomposes to sulfer dioxide and oxygen gas?
2SO2 + O2 ---> 2SO3
delta H = 198kJ
The reaction as written gives dela H = 198 kJ for 2 mols SO2 to 2 mols SO3. You want the reverse reaction so that will be -198 kJ for 2 mols SO3 to 2 mols SO2. So wouldn't 4 mols SO3 to 4 mols SO2 be twice that?
To calculate the change in enthalpy (ΔH) for a chemical reaction, you can use the stoichiometry of the reaction and the given values. Here's how you can determine the change in enthalpy for the given reaction:
1. Start by identifying the stoichiometric coefficients in the balanced chemical equation:
2SO2 + O2 → 2SO3
2. Look at the coefficients that correspond to the reactant and product of interest. In this case, you're interested in the decomposition of sulfur trioxide:
2SO2 + O2 → 2SO3
3. The coefficient of sulfur trioxide (SO3) is 2. This means that for every 2 moles of sulfur trioxide that decomposes, the enthalpy change is 198 kJ.
4. Since the question asks about 4.00 moles of sulfur trioxide decomposing, you need to determine how many times the reaction occurs. Divide the moles of sulfur trioxide by the stoichiometric coefficient:
4.00 mol SO3 ÷ 2 mol SO3 = 2.00
5. From step 4, you find that the reaction occurs 2 times for 4.00 moles of sulfur trioxide.
6. Multiply the enthalpy change by the number of times the reaction occurs:
198 kJ/mol SO3 × 2 mol SO3 = 396 kJ
Therefore, the change in enthalpy when 4.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas is 396 kJ.