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What is the change in enthalpy when 9.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas

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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.

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