Water and Ice

Chemistry Level pending

The picture above illustrates the process of water frozen into a solid state, ice. Which of the following statements is correct? ( Δ H \Delta H , Δ S \Delta S and Δ G \Delta G mean the change in enthalpy, entropy and Gibbs free energy, respectively.)

a) Both the enthalpy and entropy of the system decrease.
b) If the temperature is lower than 0 C 0^\circ\text{C} , then ( Δ G < 0 \Delta G < 0 ).
c) If the temperature is higher than 0 C 0^\circ\text{C} , then Δ H > T Δ S \lvert{\Delta H}\rvert > \lvert{T\Delta S}\rvert .

a) and b) b) and c) a) and c) a), b), and c)

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2 solutions

Lu Chee Ket
Jan 29, 2016

a) Water released energy and formed ice while water molecules become highly ordered; this makes the sentence true;

b) When temperature is lower than 0 , 0^\circ, Δ G < 0 \Delta G < 0 stands for less useful ice; only when the ambient temperature is higher than 0 , 0^\circ, the ice has potential to do work; this makes the statements true;

c) While Δ G > 0 \Delta G > 0 at temperature higher than 0 0^\circ to agree as the same as above, with Δ G = Δ H T Δ S , \Delta G = \Delta H - T \Delta S, Δ H > T Δ S \Delta H > T \Delta S does n o t not imply Δ H > T Δ S . | \Delta H | > | T \Delta S |. This could mean ice absorbs heat for positive Δ H \Delta H to do work by the system itself, but releases plenty of heat may not allow the system itself to contribute work; Δ H > T Δ S | \Delta H | > | T \Delta S | could turn out to be false always when overall disorder represented by magnitude ought not be smaller than Δ H | \Delta H | . We can't enjoy having more ice spontaneously while getting an overall output to arrange for less disorder! Quality of energy can only reduce but not increase.

Answer: a ) a n d b ) \boxed{a)~and~b)}

Jan Usman
Mar 8, 2014

v

Dude, any solution plz

Amey Meher - 7 years, 2 months ago

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