Question

Calculate the standard entropy change for the industrial synthesis of urea (a common fertilizer): CO2(g) + 2 NH3(g) → CO(NH2)2(s) + H2O(ℓ) at 25◦C. patel (smp4358) – 28: Fall Exam Review Part Three – lyssy – (112476) 8 S ◦ CO2(g) 213.74 J K·mol NH3(g) 192.45 J K·mol CO(NH2)2(s) 104.6 J K·mol H2O(ℓ) 69.91 J K·mol Answer in units of J K · mol.

Answer 1

Answer : The standard entropy change is, -424.1 J/K

Explanation :

The given balanced reaction is,

$CO_2(g)+2NH_3(g)\rightarrow CO(NH_2)_2(s)+H_2O(l)$

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

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

$\Delta S^o=[n_{CO(NH_2)_2}\times \Delta S^0_{(CO(NH_2)_2)}+n_{H_2O}\times \Delta S^0_{(H_2O)}]-[n_{CO_2}\times \Delta S^0_{(CO_2)}+n_{NH_3}\times \Delta S^0_{(NH_3)}]$

where,

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

n = number of moles

$\Delta S^0$ = standard entropy change

$\Delta S^0_{(CO(NH_2)_2)}$ = 104.6 J/mol.K

$\Delta S^0_{(H_2O)}$ = 69.91 J/mol.K

$\Delta S^0_{(CO_2)}$ = 213.74 J/mol.K

$\Delta S^0_{(NH_3)}$ = 192.45 J/mol.K

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

$\Delta S^o=[1mole\times (104.6J/K.mole)+1mole\times (69.91J/K.mole)}]-[1mole\times (213.74J/K.mole)+2mole\times (192.45J/K.mole)]$

$\Delta S^o=-424.1J/K$

Therefore, the standard entropy change is, -424.1 J/K