Sulphur Reacts with oxygen to form sulfur trioxide according to the equation :
2S+3O2--------2SO3
How many moles of sulphur React with 9 mol O2?
Answers
Answer:
Start by taking a look at the thermochemical equation given to you
2
SO
2(g]
+
O
2(g]
→
2
SO
3(g]
Δ
H
∘
rxn
=
−
198 kJ/mol
Two important things to notice here
the balanced chemical equation stipulates the formation of two moles of sulfur trioxide when two moles of sulfur dioxide and one mole of oxygen react
the standard enthalpy change of reaction carries a negative sign
The fact that the standard enthalpy change of reaction carries a negative sign tells you that
198 kJ
heat is being given off when two moles of sulfur trioxide are formed by the reaction.
Simply put, the standard enthalpy change of reaction given to you corresponds to the formation of two moles of sulfur trioxide.
The first thing to do now is determine how much heat is given off when one mole of sulfur trioxide is formed by this reaction
1
mole SO
3
⋅
198 kJ
2
moles SO
3
=
99 kJ
This tells you that the reaction
SO
2(g]
+
1
2
O
2(g]
→
SO
3(g]
will have a standard enthalpy change of reaction equal to
ΔH∘for w=−99 kJ .
Since the formation of sulfur trioxide is exothermic process, i.e. it gives off heat, the decomposition of sulfur trioxide will be an endothermic process, i.e. it will require heat.
This means that you need to change the sign of the standard enthalpy change of reaction from negative to positive. Therefore, this reaction
Explanation:
6 mole of sulfur react with 9 mole oxygen