Predict the sign of δs°for the following reaction.
cu2o(s) + c(s) → 2cu(s) + co(g)
δs° ≈ 0
δ s°< 0
δ s°> 0
more information is needed to make a reasonable prediction.

Δ S° > 0

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 the change in 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 .

From the reaction given in the question ,

Cu₂O (s)  + C (s) → 2 Cu (s) + CO (g)

Considering the states of the reactant and the product ..

The product formed i.e. CO is in gaseous species , hence , the entropy increases .

you did not give any options thus this is the best answer i can give to you with this little information you have given

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