Predict whether δs° is greater than, less than, or approximately zero for each of the following reactions, and explain your choice. (a) 4 nh3(g) + 5 o2(g) → 4 no(g) + 6 h2o(g) (b) ch4(g) + 2 o2(g) → co2(g) + 2 h2o(g) (c) caco3(s) → cao(s) + co2(g)

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase .  

As the molecules in liquid state are loosely packed and has less force of attraction between the molecules, but as it is converted to solid, the force of attraction between the molecule increases and hence entropy decreases.

From the question ,

( a )  NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

Gaseous atoms -

Reactant - 1 + 5 = 6

Product - 4 + 6 = 10 ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

( b ) CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

Gaseous atoms -

Reactant - 1 + 2 = 3

Product - 1 + 2 = 3 ,

Since ,

Both the side the value of gaseous atoms are , hence , Δ°S = 0 .

( c ) CaCO₃(s) → CaO(s) + CO₂(g)

Gaseous atoms -

Reactant = 0

Product - 0 + 1 = 1 ,

Since ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

h let me think

output?

explanation: