Explane entropy and enthalpy in detail?

Typically, enthalpy and entropy are two state functions which are known to jointly contribute to the spontaneity of a reaction.

Disclaimer: long answer! You wanted it "in detail"! :)

In general, they can be defined as follows:

ENTHALPY

Enthalpy is represented as #H#. A thermodynamics relation for it is:

#DeltaH = DeltaU + Delta(PV)#
#= q + w + PDeltaV + VDeltaP#
#= q - PDeltaV + PDeltaV + VDeltaP#
#= q + VDeltaP#

where #U# is the internal energy due to heat flow #q# and expansion work #w#, #P# is pressure, and #V# is volume.

So, enthalpy is the heat flow in a system plus that due to any change in the pressure. We typically see situations of constant pressure at a constant altitude, and so:

#DeltaH = q_p#

...so in common situations that you typically work with in General Chemistry Thermodynamics problems (constant pressure), enthalpy can be considered to just be the heat flow.

ENTROPY

Entropy is typically known to be a measure of disorder in a reaction relative to absolute zero, #"0 K"#, and is represented as #S# in #"J/K"#. Entropy can be written as:

#DeltaS >= q/T = q_"rev"/T >= 0#

where #q_"rev"# is a reversible heat flow, and is the maximum heat flow, whereas #q# in general is smaller than #q_"rev"#. So, entropy is the measure of heat flow per #"K"# or #""^o"C"#; in other words, the amount of motion one can acquire due to a temperature change. (The smaller the value, the more ordered.)

You can see that if we take this relationship and work with it, we get:

#TDeltaS = q_"rev"#

At a constant pressure, we get:

#TDeltaS = q_p = DeltaH#

and so

#DeltaG = DeltaH - TDeltaS = 0#

which is a situation that occurs when a system is at equilibrium, such as a phase change (phase changes are reversible).

This is enough for this topic.

Past here is a long example for those curious folks who want to work with enthalpy or entropy's properties as state functions.

----- LONG EXAMPLE -----

Now, a property common to both of these is that they're both state functions. You can do the following example with either entropy or enthalpy. If, let's say, you wanted the enthalpy of a reaction at #"500 K"# and you had the value for it at #"298 K"#, and you know the molar heat capacity #barC_p# of each substance, you can get that enthalpy at #"500 K"# like so:

1. Construct a path

2. Write out the calculations that allow you to go from one step to the next

3. Evaluate them

Let's say we had the following reaction at constant pressure (subscript #"p"#):

#3H_2 + N_2 rightleftharpoons 2NH_3#

Acquired from wikipedia:

#H_2(g)#: #barC_p ~~ "28.836 J/mol"*"K"#
#N_2(g)#: #barC_p ~~ "29.124 J/mol"*"K"#
#NH_3(g)#: #barC_p ~~ "35.06 J/mol"*"K"#

The next thing we need is the enthalpy of reaction at #"298 K"#, which is #-"92.38 kJ/mol"#. Now, the last thing we need is a path to use our numbers. This is a path we can use, called the Thermodynamic Cycle:

1.

We don't know the direct path to get to the goal, so we break it down into realistic steps, knowing that as a state function, enthalpy will allow us to do this.

So, we end up doing:

2.

#color(blue)(DeltaH_"rxn"^(500 K))#

#= int_(500 K)^(298 K) (barC_(p,H_2) + barC_(p,N_2))dT + DeltaH_"rxn"^o + int_(298K)^(500K) barC_(p,NH_3)dT#

3.

Treating the heat capacities as basically constants for simplicity, we just get #int_(T_1)^(T_2) barC_pdT = [barC_pT]|_(T_1)^(T_2)# or #barC_pDeltaT#:

#= (barC_(p,H_2) + barC_(p,N_2))DeltaT + DeltaH_"rxn"^o + barC_(p,NH_3)DeltaT#

#= (overbrace("28.836 J/mol"cdot"K")^("H"_2) + overbrace("29.124 J/mol"cdot"K")^("N"_2))("298 K" - "500 K") overbrace(- 9.238 xx 10^4 "J/mol")^("Known enthalpy at 298 K") + (overbrace("35.06 J/mol"*"K")^("NH"_3))("500 K" - "298 K")#

#= (-11707.92 - 9.238 xx 10^4 + 7082.12) "J/mol"#

#= -"97005.8 J/mol" ~~ color(blue)(-"97.006 kJ/mol")#

So the enthalpy at #"500 K"# is more exothermic than the one at room temperature for this reaction. This makes sense since the higher the temperature, the more heat is available to catalyze this reaction, and the more heat is released for the reaction.

If you wanted to do this for entropy, use #barC_p/T# instead of #barC_p#, and you get:

#int_(T_1)^(T_2) barC_p/TdT = barCpln|T_2/T_1|#