Question

Use the standard reaction enthalpies given below to determine Delta H_rxn degree for the following reaction: 2 S(s) + 3 O_2(g) rightarrow 2SO_3(g) Delta H_rxn degree Given: SO_2(g) rightarrow S(s) + O_2(g) Delta H_rxn degree = +296.8 kJ 2SO_2(g) + O_2(g) rightarrow 2 SO_3(g) Delta H_rxn degree = -197.8 kJ Name those compounds? Li_2SO_3 Al_2S_3 P_2O_3 COBr_2-6H_2O

Answer 1

9.

Here Hess law has to use

Hess's law states that the change of enthalpy in a chemical reaction (i.e. the heat of reaction at constant pressure) is independent of the pathway between the initial and final states.

In other words, if a chemical change takes place by several different routes, the overall enthalpy change is the same, regardless of the route by which the chemical change occurs (provided the initial and final condition are the same).

We have

SO2(g) === > S(s) + O2(g) delta H⁰rxn = +296.8 KJ ------eqn 1

And

2SO2(g) + O2(g) === > 2SO3(g) = -197.8 KJ ----------- eqn 2

Rearrange the equation (1): SO2(g) === > S(s) + O2(g) delta H⁰rxn = +296.8 KJ

as

2 [S(s) + O2(g) === > SO2(g)] = -2(296.8KJ) -- [here delta H⁰rxn became negative because of the rearrangement of equation) ----(eqn 3)

2SO2(g) + O2(g) === > 2SO3(g) = -197.8 KJ -------(eqn 4)

From equations (3 ) and (4)

SO2 will cancel out and can write the equation as

2S(s) + 3 O2(g) === > 2SO3(g)    delta H⁰rxn = -791.4 kJ

So the delta H⁰rxn for the reaction is -791.4 kJ

b)

Li2SO3 – Lithium sulphite- Lithium sulfite is prepared by treating lithium with sulfur.

Al2S3- Aluminium sulphide or aluminiumsulfide - It is formed by an exothermic reaction between aluminum and sulfur.

P2O5 is phosphorous pentoxide- It can be prepared by burning phosphorus in very dry air.

CaBr2.6H2O is calcium bromide hexahydrate