Question

Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3 = 192.5 J/mol K H2 = 130.6 J/mol K N2 = 191.5 J/mol K

Answer 1

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$.

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

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

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

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

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

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K