Which of the following reactions results in a decrease in entropy? (2 points) question 17 options: 1) 6co2 (g) + 6h2o (l) yields c6h12o6 (s) + 6o2 (g) 2) 2nh3 (g) yields n2 (g) + 3h2 (g) 3) n2o4 (g) yields 2no2 (g) 4) 2h2o2 (aq) yields 2h2o (l) + o2 (g)

Correct Option A i.e. 6CO2 (g) + 6H2O (l)→C6H12O6 (s)+ 6O2 (g)

Reason:

Entropy is the measure of degree of disorder in a system. Greater the disorder, more is the entropy.

Change in entropy of system can be predicted on the basis of change in number of moles of reactant and product. 

For instance, if Δn = ∑n products - ∑n reactants > o, it will increase entropy.
On other hand, if Δn = ∑n products - ∑n reactants < o, it will decrease entropy.

Now consider following cases
Option 1: 6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
Here, Δn = 7 - 12 = -5 . Thus, entropy of system will decrease.

Option 2: 2NH3 (g) → N2 (g) + 3H2 (g)
Here, Δn = 4 - 2 = 2 . Thus, entropy of system will increase.

Option 3: N2O4 (g) → 2NO2 (g)
Here, Δn = 2  - 1 = 1 . Thus, entropy of system will increase.

option 4: 2H2O2 (aq) → 2H2O (l) + O2 (g)
Here, Δn = 3  - 2 = 1 . Thus, entropy of system will increase.

Only one reactant yields 2 or more products. meaning you will only have 1 product and it will break down into 2 or more products. eg. h2o -> h2 + o2