Question

Consider the reaction N2(g) + O2(g) --> 2NO(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.38 moles of N2(g) react at standard conditions. S°surroundings = ___J/K

Answer 1

Enthalpy change of the reaction at 298 K.

∆Hrxn = 90.25 KJ/mol.

Now
Entropy of surroundings (∆S° surrounding)= -∆H/ T
∆S° surrounding= -90.25 / 298 KJ/mol.K
∆S° surrounding= - 0.302 KJ/mol.K
∆S° surrounding = - 302.85 J/ mol.K

Moles of N2 reacted = 2.38 mol
In the balanced reaction.
Mole ratio of N2 and NO = 1 : 2
Hence, moles of NO formed = 2 × 2.38 = 4.76 mol
So, entropy change of surroundings in 4.76 moles of NO

∆S°surroundings = 4.76× (- 302.85 ) J/K
∆S°surroundings = -1441.57 J/K.