Question

Consider the reaction: H2(g) + Cl2(g) --> 2HCl(g)

Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.28 moles of H2(g)react at standard conditions.

Answer 1

Answer : The entropy change of reaction for 2.28 moles of
`H_2` reacts at standard condition is 45.8 J/K

Explanation :

The given balanced reaction is,

`H_2(g)+Cl_2(g)\rightarrow 2HCl(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_(HCl)* \Delta S_f^0_((HCl))]-[n_(H_2)* \Delta S_f^0_((H_2))+n_(Cl_2)* \Delta S_f^0_((Cl_2))]`

where,

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

n = number of moles

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

`\Delta S_f^0_((H_2))` = 130.684 J/mol.K

`\Delta S_f^0_((Cl_2))` = 223.066 J/mol.K

`\Delta S_f^0_((HCl))` = 186.908 J/mol.K

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

`\Delta S^o=[2mole* (186.908J/K.mole)]-[1mole* (130.684J/K.mole)+1mole* (223.066J/K.mole)}]`

`\Delta S^o=20.066J/K`

Now we have to calculate the entropy change of reaction for 2.28 moles of
`H_2` reacts at standard condition.

From the reaction we conclude that,

As, 1 moles of
`H_2` has entropy change = 20.066 J/K

So, 2.28 moles of
`H_2` has entropy change =
`(2.28)/(1)* 20.066=45.8J/K`

Therefore, the entropy change of reaction for 2.28 moles of
`H_2` reacts at standard condition is 45.8 J/K