Question

Sulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆Hr = –296.5 kJ mol–1
2S(s) + 3O2(g) → 2SO3(g) ∆Hr = –791.4 kJ mol–1
Sulfur trioxide can be made from sulfur dioxide and oxygen.
2SO2(g) + O2(g) → 2SO3(g)
What is the standard enthalpy change for this reaction? PLEASE I WANT THE METHOD..

Answer 1

Answer:

The standard enthalpy change of given reaction will be $-198.4kJmol^{-1}$

Explanation:

We will use Hess's law approach to solve the problem.

According to Hess's law the enthalpy of a reaction depends upon only on the initial and final state.

It is independent of the path taken by the reaction.

So we may combine the two given equations to obtain the third equation.

The enthalpy of third equation will be the arithmetic combination of the two enthalpies.

$S(s)+O_{2}(g)--->SO_{2}(g) \Delta Hr=-296.5 kJmol^{-1}............(1)$

$2S(s)+3O_{2}(g)--->2SO_{3}(g) \Delta Hr = -791.4 kJmol^{-1} ........(2)$

$2SO_{2}(g)+O_{2}(g)--->2SO_{3}(g) \Delta Hr = ??............(3)$

Equation three can be obtained by:

$equation(3)=equation(2)-[2X(equation(1)]$

$\Delta H_{eq3}=\Delta H_{eq2}-[2X(\Delta H_{eq1})]\\\Delta H_{eq3}=-791.4-[2X(-296.5)]=-198.4kJmol^{-1}$