Book says -

"(f) Enthalpy of Dilution

It is known that enthalpy of solution is the enthalpy change associated with the addition of a specified amount of solute to the specified amount of solvent at a constant temperature and pressure. This argument can be applied to any solvent with slight modification. Enthalpy change for dissolving one mole of gaseous hydrogen chloride in 10 mol of water can be represented by the following equation. For convenience we will use the symbol aq. for water

HCl(g) + 10 aq. → HCl.10 aq.

∆H = –69.01 kJ / mol

Let us consider the following set of enthalpy changes:

(S-1) HCl(g) + 25 aq. → HCl.25 aq.

∆H = –72.03 kJ / mol

(S-2) HCl(g) + 40 aq. → HCl.40 aq.

∆H = –72.79 kJ / mol

(S-3) HCl(g) + ∞ aq. → HCl. ∞ aq.

∆H = –74.85 kJ / mol

The values of ∆H show general dependence of the enthalpy of solution on amount of solvent. As more and more solvent is used, the enthalpy of solution approaches a limiting value, i.e, the value in infinitely dilute solution. For hydrochloric acid this value of ∆H is given above in equation (S-3).

If we subtract the first equation (equation S-1) from the second equation (equation S-2) in the above set of equations, we obtain–

HCl.25 aq. + 15 aq. → HCl.40 aq.

∆H = [ –72.79 – (–72.03)] kJ / mol = – 0.76 kJ / mol

This value (–0.76kJ/mol) of ∆H is enthalpy of dilution. It is the heat withdrawn from the surroundings when additional solvent is added to the solution. The enthalpy of dilution of a solution is dependent on the original concentration of the solution and the amount of solvent added"

Question

1. What is the convention here I get the idea of dilution. Why it says to substract then goes on to add one side and subtract the other ?

2. "It is the heat withdrawn from the surroundings when additional solvent is added to the solution." - the value is negative that means heat is given out ?

2 hours ago, HbWhi5F said:

Book says -

"(f) Enthalpy of Dilution

It is known that enthalpy of solution is the enthalpy change associated with the addition of a specified amount of solute to the specified amount of solvent at a constant temperature and pressure. This argument can be applied to any solvent with slight modification. Enthalpy change for dissolving one mole of gaseous hydrogen chloride in 10 mol of water can be represented by the following equation. For convenience we will use the symbol aq. for water

HCl(g) + 10 aq. → HCl.10 aq.

∆H = –69.01 kJ / mol

Let us consider the following set of enthalpy changes:

(S-1) HCl(g) + 25 aq. → HCl.25 aq.

∆H = –72.03 kJ / mol

(S-2) HCl(g) + 40 aq. → HCl.40 aq.

∆H = –72.79 kJ / mol

(S-3) HCl(g) + ∞ aq. → HCl. ∞ aq.

∆H = –74.85 kJ / mol

The values of ∆H show general dependence of the enthalpy of solution on amount of solvent. As more and more solvent is used, the enthalpy of solution approaches a limiting value, i.e, the value in infinitely dilute solution. For hydrochloric acid this value of ∆H is given above in equation (S-3).

If we subtract the first equation (equation S-1) from the second equation (equation S-2) in the above set of equations, we obtain–

HCl.25 aq. + 15 aq. → HCl.40 aq.

∆H = [ –72.79 – (–72.03)] kJ / mol = – 0.76 kJ / mol

This value (–0.76kJ/mol) of ∆H is enthalpy of dilution. It is the heat withdrawn from the surroundings when additional solvent is added to the solution. The enthalpy of dilution of a solution is dependent on the original concentration of the solution and the amount of solvent added"

Question

1. What is the convention here I get the idea of dilution. Why it says to substract then goes on to add one side and subtract the other ?

2. "It is the heat withdrawn from the surroundings when additional solvent is added to the solution." - the value is negative that means heat is given out ?

Re (2), yes.

As was explained on one of your previous threads, the convention is that energy lost from the system is -ve and energy added to it is +ve. So an exothermic process will have a -ve enthalpy change (ΔH<0). In this case the more you dilute HCL the more energy is released (it gets warm), so the change is -ve.

Yr (1) involves calculating the difference between 2 -ve quantities, that's all. Don't get bamboozled by all the -ve signs.

Edited September 7Sep 7 by exchemist

@exchemist why is it negative and book says process is endothermic ?

34 minutes ago, HbWhi5F said:

@exchemist why is it negative and book says process is endothermic ?

What book is this?

Edited September 8Sep 8 by exchemist

9 hours ago, exchemist said:

As was explained on one of your previous threads, the convention is that energy lost from the system is -ve and energy added to it is +ve.

One thing I'd like to point out to @HbWhi5F is that everything in thermodynamics is about the system and everything that happens to the system. For example, suppose one has some gas in a cylinder with a piston, and one presses on the piston to compress the gas. The work done on the gas depends on the pressure of the gas inside the cylinder and the decrease in volume of the gas inside the cylinder. All well and good. But now release the piston and allow the gas to expand. The work done by the gas now depends on the pressure of the environment. That's no good because the environment is outside of the system which everything is about. So what one has to do is make the pressure of the environment equal to the pressure inside the cylinder at all times. But if the two pressures are exactly equal, there will be no compression or expansion of the gas. So to make the gas compress, one makes the pressure outside the cylinder infinitesimally greater than the pressure inside the cylinder; and to make the gas expand, one makes the pressure outside the cylinder infinitesimally less than the pressure inside the cylinder. Because the two pressures differ by only an infinitesimal amount, they can be regarded as equal. Now the work done by the expanding gas depends on the pressure inside the cylinder (because it is equal to the pressure outside the cylinder) and the increase in volume of the gas inside the cylinder. Now everything is about the system.

Also, note that expansion and compression of the gas inside the cylinder are the reverse of each other. Also note that to go from compression to expansion (or from expansion to compression), one only needs to change the pressure outside the cylinder from infinitesimally greater than to infinitesimally less than the pressure inside the cylinder (or from infinitesimally less than to infinitesimally greater than the pressure inside the cylinder). In either case, only an infinitesimal change in pressure is required to reverse the process. Processes that are reversed by applying an infinitesimal change are called reversible processes. Reversible processes are very important in thermodynamics because thermodynamic quantities such as enthalpy and entropy are defined in terms of them. You may have noticed that only in the case of the reversible compression and expansion of the gas inside the cylinder that you get back all of the work that you put into it. In the case of the irreversible expansion, because the pressure outside the cylinder must be less than the pressure inside the cylinder (in order for the gas to expand), the work done by the expanding gas must be less than the work done on the gas when it was compressed.

So, what happens to the compression work not done by the gas during its irreversible expansion? It gets converted to heat. The heat increases the temperature of the gas inside the cylinder. Now let the temperature inside the cylinder cool to the temperature it was before it was compressed. The state of the system (its pressure, volume, and temperature) is now the same as it was before the compression. But the total work of the system is not the same. Neither is the total heat of the system. That is, by performing some process of changing the state of the system and eventually returning the system to the original state, neither the work nor the heat of the system return to the same value. It is said that neither work nor heat are functions of state. A function of state is a function that returns to the same value when the state variables return to the same values after traversing any path in the space of states. However, reversible processes do ensure that quantities do return to the same value after the system returns to the same state. Thus, thermodynamic quantities such as enthalpy and entropy are functions of state.