Question

What is the heat of reaction when sulfur dioxide reacts with oxygen to form sulfur trioxide? 2so2(g) + o2(g) → 2so3(g) (∆hf0 so2(g) = –298.8 kj/mol; ∆hf0 so3(g) = –395.7 kj/mol) –201.9 kj –492.6 kj –694.5 kj –193.8 kj?

Answer 1

Answer : The enthalpy change for this reaction is -193.8 kJ.

Solution :

The balanced chemical reaction is,

`2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)`

The expression for enthalpy change is,

`\Delta H=\sum [n* \Delta H_f(product)]-\sum [n* \Delta H_f(reactant)]`

`\Delta H=[(n_(SO_3)* \Delta H_(SO_3))]-[(n_(O_2)* \Delta H_(O_2))+(n_(SO_2)* \Delta H_(SO_2))]`

where,

n = number of moles

`\Delta H_(O_2)=0` (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

`\Delta H=[(2* -395.7)]-[(1* 0)+(2* -298.8)]`

`Delta H=-193.8kJ`

Therefore, the enthalpy change for this reaction is, -193.8 KJ