Question

Consider the reaction: 2BrF3(g) --> Br2(g) + 3F2(g)

Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.62 moles of BrF3(g)react at standard conditions.

Answer 1

Answer : The entropy change of reaction for 1.62 moles of
`BrF_3` reacts at standard condition is 217.68 J/K

Explanation :

The given balanced reaction is,

`2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)`

The expression used for entropy change of reaction
`(\Delta S^o)` is:

`\Delta S^o=S_f_(product)-S_f_(reactant)`

`\Delta S^o=[n_(Br_2)* \Delta S_f^0_((Br_2))+n_(F_2)* \Delta S_f^0_((F_2))]-[n_(BrF_3)* \Delta S_f^0_((BrF_3))]`

where,

`\Delta S^o` = entropy change of reaction = ?

n = number of moles

`\Delta S_f^0` = standard entropy of formation

`\Delta S_f^0_((Br_2))` = 245.463 J/mol.K

`\Delta S_f^0_((F_2))` = 202.78 J/mol.K

`\Delta S_f^0_((BrF_3))` = 292.53 J/mol.K

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

`\Delta S^o=[1mole* (245.463J/K.mole)+3mole* (202.78J/K.mole)}]-[2mole* (292.53J/K.mole)]`

`\Delta S^o=268.74J/K`

Now we have to calculate the entropy change of reaction for 1.62 moles of
`BrF_3` reacts at standard condition.

From the reaction we conclude that,

As, 2 moles of
`BrF_3` has entropy change = 268.74 J/K

So, 1.62 moles of
`BrF_3` has entropy change =
`(1.62)/(2)* 268.74=217.68J/K`

Therefore, the entropy change of reaction for 1.62 moles of
`BrF_3` reacts at standard condition is 217.68 J/K