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,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{SO_3}\times \Delta H_{SO_3})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{SO_2}\times \Delta H_{SO_2})]$

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

$\Delta H=[(2\times -395.7)]-[(1\times 0)+(2\times -298.8)]$

$Delta H=-193.8kJ$

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