Chapter 51: Problem 3

Consider the freezing of water: \(\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\). a) Will \(\Delta H\) be positive or negative? b) Will \(\Delta S\) be positive or negative? c) Explain your reasoning for both of your answers.

Short Answer

Expert verified

a) The ∆H will be negative as heat is released during freezing; b) The ∆S will be negative as freezing decreases the disorder of the system.

Step by step solution

Analyzing Change in Enthalpy (∆H)

Enthalpy quantifies the total heat content of a system. Any exothermic reaction, like freezing, releases heat. So for the reaction \( \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \), ∆H would be negative, since the system is releasing heat to its surroundings.

Analyzing Change in Entropy (∆S)

Entropy quantifies the degree of randomness or disorder in a system. The conversion from a liquid state to a solid state reduces the randomness of water molecules. They move from a fairly disordered state (with relatively free movement) to a highly ordered, fixed lattice structure. Thus, the entropy of the system decreases during freezing, meaning ∆S is negative.

Recapitulation

In the case of water freezing, because heat is being released and the disorder is decreasing, both ∆H and ∆S should be negative.

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Consider the freezing of water: \(\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\). a) Will \(\Delta H\) be positive or negative? b) Will \(\Delta S\) be positive or negative? c) Explain your reasoning for both of your answers.

Short Answer

Step by step solution

Analyzing Change in Enthalpy (∆H)

Analyzing Change in Entropy (∆S)

Recapitulation

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