Question

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

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:

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]$

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:

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

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

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 = $\frac{1.62}{2}\times 268.74=217.68J/K$

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