Question

For which of the following processes will \DeltaΔS be negative?PbCl2(s) = Pb2+(aq) + 2 Cl-(aq)MgO(s) + CO2(g) = MgCO3(s)CO2(aq) = CO2(g)C5H12(l) + 8 O2(g) = 5 CO2(g) + 6 H2O(g)

Answer 1

Answer

a) $PbCl_2(s)\rightarrow Pb^{2+}(aq)+2Cl^-(aq)$: $\Delta S$ = +ve

b) $MgO(s)+CO_2(g)\rightarrow MgCO_3(s)$: $\Delta S$ = -ve

c) $CO_2(aq)\rightarrow CO_2(g)$: $\Delta S$= +ve

d)  $C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)$: $\Delta S$ = +ve

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.

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

As solid is moving to ions form , randomness increases and thus sign of $\Delta S$ is positive.

b)$MgO(s)+CO_2(g)\rightarrow MgCO_3(s)$

As gaseous reactants are converted to solid products , randomness decreases and thus sign of $\Delta S$  is negative.

c) $CO_2(aq)\rightarrow CO_2(g)$

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

d) $C_5H_{12}(l)+8O_2(g)rightarrow 5CO_2(g)+6H_2O(g)$

As 8 moles of gaseous reactants are converted to 11 moles of gaseous products , randomness increases and thus sign of $\Delta S$ is positive.