Question

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

Answer 1

292 kJ
Explanation: the summ of the first and the second reaction.

Answer 2

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