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Bogdan [553]
3 years ago
7

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)
Chemistry
1 answer:
mr Goodwill [35]3 years ago
7 0

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.

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