# Why is the melting of ice a spontaneous reaction?

So here's my question: we know that melting of ice is an spontaneous reaction but is it spontaneous? We have to supply energy in the form heat for the reaction to take place.

First of all,

what exactly is spontaneous reaction?

In very easy language, reaction that occurs in a given set of conditions without intervention is called spontaneous reaction.

Now let us consider melting of ice example,

Take a ice on a plate and leave it for half an hour(conditions). After half an hour we will notice water instead of ice.

This process took place without any intervention from your side. Heat absorbed by the ice from surrounding is due to set of conditions not due to your intervention.

Thus we can consider melting of ice as a spontaneous process(or reaction).

I'm not sure if you have learnt this, so I just tell the story from the beginning. (I'm not sure, but if this is hard to understand, please tell me.)

There is a universal law usually called as the second law of thermodynamics. It states:

The entropy of the universe tends to a maximum. If a process is reversible, the entropy of the universe will keep unchanged; if it is irreversible, the entropy increases.

Well, it's not the original statement. Historically, from the original statement of the law, a thermodynamic function is considered useful; that's "entropy" I mentioned here.

So what is entropy? It's quite a mystery because at the first set out people do not defined what will it be; while, they just defined how it changes.

An infinitesimal increment in the entropy of a system results from an infinitesimal transfer of heat is $\mathrm{d}S=\delta Q/T$

Ur, I think you will not like this. Hopefully, later, Boltzmann state a hypothesis, he assumed the occupation of any microstate (means the distribution of all the particles we concerned in the space we concerned) is equally probable, then

In most cases, the entropy is proportional to the logarithm of the number of possible microstate.

You propably still think this is hard to understand... so we can say all the above in a easier way (but, I think, a less accurate way):

the second law of thermodynamics teaches us, the randomness of the universe tend to be maximum. If a process is reversible, the randomness of the universe will keep unchanged; if it is irreversible, the randomness increases.

But...we still can't know how to judge whether a process is spontaneous or not yet. However, by the second law of thermodynamics we are easy to know the randomness of universe increases in a spontaneous process.

Thus we get $\Delta S_\text{universe} = \Delta S_\text{system} + \Delta S_\text{surronding} > 0$.

While we also know if the heat transferred to the system from the surroundings is $Q$, then $\Delta S_\text{surronding} = -Q/T$, where $T$ is the temperature of the surrounding. (Since the surrounding is large, we may assume the temperature is unchanged.)

So we get $\Delta S_\text{system} - Q/T > 0$, i.e. $T\Delta S_\text{system} - Q > 0$

if the process is now assumed to be isobaric (means that the barometric pressure of system is unchanged), then $Q = \Delta H$ and we get $T\Delta S_\text{system} - \Delta H > 0$.

Now we can define a new thermodynamic function:

The Gibbs free energy of a system is $G = H - TS$.

Then we can say that in a spontaneous process, $\Delta G < 0$. (If $\Delta G = 0$, it is in equilibrium.)

Maybe you have known it all... uh... so let's solve the question: Why is the melting of ice a spontaneous reaction?

You know in liquid particles will arrange in a messer way then in a solid, so the entropy, or the randonness of the "reaction system" is increased, i.e. $\Delta S_\text{system} > 0$. However, when ice melt, $\Delta H > 0$, so for $\Delta G < 0$ you need to keep the temperature higher, or supply energy to it. And simliarly, you will find if temperature is low enough, the water will spontaenously freeze. This seems to be the case you are familiar with.

Noticed that sometimes $\Delta G < 0$, however the speed of reaction is too low, so we supply energy for accerlation. For example, the diamond will burn: $\ce{C_{(s, diamond)} + O2_{(g)} -> CO2_{(g)}},\ \Delta G^\circ= -397\ \mathrm{kJ}$.

But you never see this happen in common situation: because it's too slow.

• – Gowtham Apr 28 '15 at 14:01
• As long as the net entropy change of system + surroundings is positive, right? – Papul May 9 '15 at 19:56

When T > 0 °C, ice melts spontaneously; the reverse process, liquid water turning into ice, is not spontaneous at these temperatures. However, when T < 0 °C, the opposite is true. Liquid water converts into ice spontaneously, and the conversion of ice into water is not spontaneous

At T = 0 °C the solid and liquid phases are in equilibrium. At this particular temperature the two phases are interconverting at the same rate, and there is no preferred direction for the process: Both the forward and reverse processes occur with equal preference, and the process is not spontaneously favored in one direction over the other.

When ΔG is negative, a process or chemical reaction proceeds spontaneously in the forward direction.

Also from the wikipedia article on Spontaneous reaction says:

Every reactant in a spontaneous process has a tendency to form the corresponding product. This tendency is related to stability. Stability is gained by a substance if it is in a minimum energy state or is in maximum randomness. Only one of these can be applied at a time. e.g. Water converting to ice is a spontaneous process because ice is more stable since it is of lower energy. However, the formation of water is also a spontaneous process as water is the more random state.

I wouldn't call it a reaction, it's a phase change. The term reaction refers to a change in covalent bonding between components of a system.

The phase change occurs in a seemingly spontaneous manner because the environment is capable of donating enough free energy (most likely from enthalpy, could happen otherwise) to overcome the energy needed to activate the process of reorganization of the system (increase in entropy). The other responses have done a nice job in terms of symbolic rigor, so I won't repeat them here.