What is the entropy change for the third reaction below? n2(g) + o2(g) → 2no(g), ∆h = 180 kj 2no2(g) → 2no(g) + o2(g), ∆h = 112 kj n2(g) + 2o2(g) → 2no2(g) a) 34 kj b) 292 kj c) 146 kj d) 68 kj

Given that  

N2(g) + O2(g) → 2NO(g), ∆H = 180 kJ  1

2NO2(g) → 2NO(g) + O2(g), ∆H = 112 kJ 2

N2(g) + 2O2(g) → 2NO2(g)    ∆H = ? kJ  3  

To calculate ∆H of equation 3 use  Hess's law reverse the equation 2 and then add it with equation 1 as follows:

N2(g) + O2(g) --> 2NO(g) ∆H = 180 kJ  

2NO(g) + O2(g --> 2NO2(g) ∆H = -112 kJ  

N2(g) + 2O2(g) --> + 2NO2(g)  ∆H = 68 kJ  

Hence the correct is D

the answer is   c.) 1575 j.

explanation: