Consider the reaction: 2N2(g) + O2(g)2N2O(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.19 moles of N2(g) react at standard conditions. S°system = J/K

ΔS = -114.296 J/K

Explanation:

Step 1: Data given

Temperature = 298 K

Number of moles N2 = 2.19 moles

S°(N2) = 191.6 J/K*mol

S°(O2)= 161.1 J/k*mol

S°(N2O) = 219.96 J/K*mol

Step 2: The balanced equation

2N2(g) + O2(g) ⇆ 2N2O(g)

Step 3: Calculate ΔSrxn

ΔSrxn = ∑S°products -  ∑S°reactants

ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol

ΔSrxn = -104.38 J/K

Step 4: Calculate ΔS for 2.19 moles

The reaction is for 2 moles N2

ΔS = -114.296 J/K

why do the isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions? chemical reactions involve changes to electrons in an atom; isotopes do not differ in electrons, and so they react the same way during chemical changes.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"