Question

Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or negativea) 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)

Answer 1

Answer: a) $Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$:  negative

b)  $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$ : positive

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$: positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)$ : negative

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$:  positive.

f) $I_2(s)\rightarrow I_2(g)$ : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

a) $Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$

As ions are moving to solid form , randomness decreases and thus sign of $\Delta S$ is negative.

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)$

As gas is changing to solid, randomness decreases and thus sign of $\Delta S$ is negative.

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

f) $I_2(s)\rightarrow I_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.