Answer:
-891kj
Step-by-step explanation:
We have the chemical equation as:
2SO2(g) + O2(g) ----2SO3; ∆H° = 198kj
If sulphur trioxide decomposes we have opposite reaction. Enthalpy of decomposition will not be positive for forward reaction.
2 moles of SO3 ----∆H° = -198Kj
1 mole = -198/2
9 moles = -198/2 x 9
= -891kj
Therefore change in enthalpy when 9 moles is decomposes is -891kj.