Given the thermochemical equation 2SO2 + O2 → 2SO3, ΔH°rxn = -198 kJ/mol, what is the standard enthalpy change for the decomposition of one mole of SO3?
99 kJ/mol
You want 1 mol SO3 and not 2 so divide by 2. You want the reverse so change the sign.
To find the standard enthalpy change for the decomposition of one mole of SO3, we need to reverse the given thermochemical equation.
The given equation is 2SO2 + O2 → 2SO3 with ΔH°rxn = -198 kJ/mol.
By reversing the equation, we get:
2SO3 → 2SO2 + O2
Now, we need to change the sign of ΔH°rxn because we reversed the equation. Therefore, the new value for ΔH°rxn is +198 kJ/mol.
This means that the standard enthalpy change for the decomposition of one mole of SO3 is +198 kJ/mol.
To find the standard enthalpy change for the decomposition of SO3, we need to use the given thermochemical equation and the concept of Hess's Law.
Hess's Law states that the standard enthalpy change of a reaction can be calculated from the standard enthalpy changes of other reactions by manipulating and combining them algebraically.
In this case, we need to reverse the given thermochemical equation to obtain the equation for the decomposition of SO3:
2SO3 → 2SO2 + O2
Since we reversed the equation, the sign of the standard enthalpy change also changes. So, the standard enthalpy change for the decomposition of SO3 is the negative of ΔH°rxn of the given equation:
ΔH°decomp = -(-198 kJ/mol) = 198 kJ/mol
Therefore, the standard enthalpy change for the decomposition of one mole of SO3 is 198 kJ/mol.