Predict whether Delta S is positive, negative, or not predictable given no furth

Question

Predict whether Delta S is positive, negative, or not predictable given no further in formation (circle the correct word) for each of the following processes, assuming each occurs at constant temperature. CaCo_3(S) rightarrow CaO(s) + CO_2(g); ..................... delta S is positive/negative/not predictable N_2(g) + 3H_2(g) rightarrow 2NH_3(g);................ Delta S is positive/negative/not predictable 3C_2H_2(g) rightarrow c_6H_6(g);.................. Delta S is positive/negative/not predictable AgCl(s) rightarrow Ag^+ (aq) + Cgamma (aq);................................. Delta S is positive/negative/not predictable

Explanation / Answer

S defines the randomness , Higher the randomness higher the value of S.

For reactions having gases, N (Number of gaseous molecules in products -Number of gaseous moles in reactants) >0 , Then randomnes increases . Hence higher S value that of reactant.

SO lets see the following reaction

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

Here we can see that CO2 is getting released , which is gas .SO from a solid reactant we have gas so randomness has increased.

here N = 1, hence randomness increases, entropy is positive

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

N = 2-4 =-2 <0 , SO entropy will decrease/negative

c)3C2H2 -> C6H6

N =1-3=-2 , So entropy will decrease/negative

d) AgCl(s) -------------->Ag+ (aq)+ Cl- (aq)

In this case there is no involvement of gases So look it in other way . Here a solid is breaking into ions and going into a solution .Hence a solution has higher randomness than a solid . SO we have positive entropy here.

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