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 d : ) have a good da

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g