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Predict the sign of the entropy change, Delta S, for each of the following reactions.?

Predict the sign of the entropy change, Delta S, for each of the following reactions.

a) Pb^2+(aq) + 2Cl-(aq) ---> PbCl2(s)

b) CaCO3(s) ---> CaO(s) + CO2 (g)

c) 2NH3(g) ---> N2(g) + 3H2(g)

d) P4(g) + 5O2(g) ---> P4O10(s)

e) C4H8(g) + 6O2(g) ---> 4CO2(g) + 4H2O(g)

f) I2(s) ---> I2(g)

The answer is either positive or negative for each...thank you soo much, I don't know how to figure this one out and I really appreciate the help!

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This is the best answer for Predict the sign of the entropy change, Delta S, for each of the following reactions.?

many particles --> fewer particles (negative entropy)

solid --> liquid --> gas (increasing entropy; more freedom of molecular movement)

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a) formation of one molecule lead chloride from two molecules, means less randomness, then there is a decrease in S; which means delta S is negative

b) formation of two molecules from one molecule of CaCo3, means more randomness, then there is an increase in S; which means delta S is positive

c) formation of four molecules from two molecule of reactants, means more randomness, then there is an increase in S; which means delta S is positive

d)formation of one molecule from four molecules of reactants, means less randomness, then there is a decrease in S; which means delta S is negative

e)  formation of eight molecules from seven molecule of reactants, means more randomness, then there is an increase in S; which means delta S is positive

f) In this case, one molecule is reactant and one is product, but the product is a gas, and gas has more randomness than solid or liquid, then there is an increase in S; which means delta S is positive.
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