# Why enthalpy of adsorption is negative?

To illustrate, let's take a hypothetical vacuum space. In that space, there is a solid and a gas, the gas is getting adsorbed over the solid. In that case, if we'll use momentum conservation, we'll get that the net kinetic energy of system has increased due to positive work done by the internal forces, but heat exchanged from system to surrounding or vice-versa is zero( as there is no outside interaction present. Therefore, dU>0 as U is sum of kinetic and potential energy of system( note that I haven't made the internal force as due to a field with a assigned potential function). We know dH=dU+PdV for processes in which the pressure outside the system is constant( which we have assumed). Therefore, as dV=0( because I had taken a fixed volume of space), therefore dH>0. But in books it is written dH due to adsorption is negative. Also, if I had assigned a potential function for internal forces, what change would have taken place? Please help me out.

• In effect, you discuss a system where gas molecules don't interact with walls except for elastic collisions. In such a system, indeed, adsorption won't be exothermic (and consequently, won't happen at all). – Ivan Neretin Apr 10 '19 at 7:53
• In what case, will it be exothermic?( I mean what kind of interaction with wall, like you specify is required)? and what is the change if I introduced potential energy for the internal forces? – MRG Apr 10 '19 at 8:05
• An attractive interaction, of course. – Ivan Neretin Apr 10 '19 at 8:06

I will point out another equation:

$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S$$

For adsorption $$\mathrm{d}S<0$$, as molecules go more organized.

Therefore also $$\mathrm{d}H < 0$$ must be true, otherwise $$\mathrm{d}G < 0$$, needed for a spontaneous process, would not be achieved.

• Its ok but my question why enthalpy is negative according to my model and your answer is justifying enthalpy must negative for adsorption being spontaneous but not how exactly it is. – MRG Apr 10 '19 at 15:16
• For the same reason like the condensation from gaseous phase, or a cation binding an electron. The energy is released by the bonding. – Poutnik Apr 10 '19 at 15:24
• You have deliberately denied for the system to be in equilibrium with the surrounding, so the dH=0 by principle. – Poutnik Apr 10 '19 at 16:02

Your mistake is in saying that the heat exchanged is zero. When we talk about a reaction being exothermic, it is based on the condition that the initial and final temperatures of the system are equal, and that, to achieve this, we exchange whatever amount of heat that is necessary.