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## How does the bond energy for the formation of #"NaCl"(s)# take place? Please explain in steps comprehensively but concisely. Thank you in advance.

Truong-Son N.
Featured 6 months ago

Do you mean the energy involved in the process to form $\text{NaCl} \left(s\right)$ in its formation reaction? Well, to form the lattice, we have to get to the ions in their gaseous state and then allow sodium to transfer its electron over to chlorine.

I get:

$\Delta {H}_{\text{lattice" -= -|DeltaH_"lattice}} |$

$= - \left[\Delta {H}_{f , \text{NaCl"(s))^@ + DeltaH_("sub","Na") + "IE"_(1,"Na"(g)) + 1/2DeltaH_("bond","Cl"_2(g)) - "EA"_(1,"Cl} \left(g\right)}\right]$,

where all the numbers you plug in to the righthand side are positive.

Formation reactions generate $\text{1 mol}$ of the product, such as $\text{NaCl} \left(s\right)$, from the individual elements ($\text{Na}$, $\text{Cl}$) in their elemental state at ${25}^{\circ} \text{C}$ and standard pressure.

(You should check your textbook to see whether it is defined as $\text{1 atm}$ or $\text{1 bar}$.)

Elemental state simply means the natural state you find them at typical life conditions.

• Sodium is a metal at room temperature and pressure.
• Chlorine is a diatomic gas at room temperature and pressure.

So, the balanced formation reaction is:

$\text{Na"(s) + 1/2"Cl"_2(g) -> "NaCl} \left(s\right)$

Note that there exists a negatively-signed lattice energy for the following reaction:

$\text{Na"^(+)(g) + "Cl"^(-)(g) -> "NaCl} \left(s\right)$

since forming the lattice stabilizes the components (we went from two gases to one solid, creating order), decreasing their overall energy.

These share the same product, so let's say we wanted to form the following cycle:

$\text{Na"(s) + 1/2"Cl"_2(g) -> [...] -> "Na"^(+)(g) + "Cl"^(-)(g) -> "NaCl} \left(s\right)$

This would be what we call the Born-Haber thermodynamic cycle. This is allowed since enthalpy is a state function, which does not depend on the path.

As it turns out, the lattice energy is difficult to acquire in real life, so it is often indirectly obtained from calculations not unlike the ones we're doing.

Our goal is to fill in the missing steps by transforming each element one step at a time (phases, charges, and chemical bonds):

${\text{Na"(s) stackrel("Sublimation")overbrace(->) "Na"(g) stackrel("1st Ionization")overbrace(->) "Na}}^{+} \left(g\right)$

$\frac{1}{2} {\text{Cl"_2(g) stackrel(1/2xx"Bond-Breaking")overbrace(->) "Cl"(g) stackrel("1st Electron Affinity")overbrace(->) "Cl}}^{-} \left(g\right)$

For simplicity in our Born-Haber cycle, let up use positive numbers and down use negative numbers.

Then, the following reactions would be involved to transform each reactant for the formation of $\boldsymbol{\text{1 mol}}$ of $\text{NaCl} \left(s\right)$ (the units therefore become $\underline{\underline{\text{kJ}}}$!):

${\text{NaCl"(s) -> "Na"(s) + 1/2"Cl}}_{2} \left(g\right)$, #-DeltaH_(f,"NaCl"(s))^@ = "411 kJ"#

$\text{Na"(s) -> "Na} \left(g\right)$, #DeltaH_("sub","Na") = "107 kJ"#

${\text{Na"(g) -> "Na}}^{+} \left(g\right) + {e}^{-}$, $\text{IE"_(1,"Na"(g)) = "502 kJ}$

$\frac{1}{2} \text{Cl"_2(g) -> "Cl} \left(g\right)$, #1/2DeltaH_("bond","Cl"_2(g)) = 1/2xx"242 kJ"#

${\text{Cl"(g) + e^(-) -> "Cl}}^{-} \left(g\right)$, $\text{EA"_(1,"Cl"(g)) = -"355 kJ}$

$\text{Na"^(+)(g) + "Cl"^(-)(g) -> "NaCl} \left(s\right)$, #DeltaH_"lattice" = ???#

$\text{-----------------------------------------------------------------------------}$

$\text{These cancel out completely upon adding, proving}$

$\text{we have a complete cycle.}$

Put that together by adding all the given energies, and you'll have the following diagram:

Take the $\Delta {H}_{f}^{\circ}$ step as being upwards to generate a complete cycle, for which $\Delta {H}_{\text{cycle}} = 0$ (since ${H}_{f} = {H}_{i}$ for a complete cycle). Therefore, we have this equation:

$0 = \Delta {H}_{\text{cycle" = DeltaH_(f,"NaCl"(s))^@ + DeltaH_("sub","Na") + "IE"_(1,"Na"(g)) + 1/2DeltaH_("bond","Cl"_2(g)) - "EA"_(1,"Cl"(g)) - DeltaH_"lattice}}$

Solving for $\Delta {H}_{\text{lattice}}$ usually gives a positive answer, so we take the negative of the answer by convention to get:

$\textcolor{b l u e}{\Delta {H}_{\text{lattice" -= -|DeltaH_"lattice}} |}$

$= \textcolor{b l u e}{- \left[\Delta {H}_{f , \text{NaCl"(s))^@ + DeltaH_("sub","Na") + "IE"_(1,"Na"(g)) + 1/2DeltaH_("bond","Cl"_2(g)) - "EA"_(1,"Cl} \left(g\right)}\right]}$

where all the numbers you plug in are positive. For example, we'd get:

$\textcolor{b l u e}{\Delta {H}_{\text{lattice","NaCl} \left(s\right)}}$

$= - \left[411 + 107 + 502 + \frac{1}{2} \left(242\right) - 355\right] \text{kJ}$

$= \textcolor{b l u e}{- \text{786 kJ}}$

## What are three colligative properties of solutions?

Truong-Son N.
Featured 3 months ago

1) The lowering of the solvent's vapor pressure.
2) The decrease in the solvent freezing point.
3) The increase in the solvent boiling point.

Heck, I could list a fourth:

4) The increase in osmotic pressure.

VAPOR PRESSURE REDUCTION

This follows from Raoult's Law for ideal solutions:

${P}_{A} = {\chi}_{A \left(v\right)} P = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*}}$

where:

• ${\chi}_{A \left(l\right)}$ is the mol fraction of the solvent $A$ in the liquid phase.
• ${\chi}_{A \left(v\right)}$ is the mol fraction of the solvent $A$ in the vapor phase.
• ${P}_{A}$ is the partial pressure of the solvent $A$ above the solution.
• ${P}_{A}^{\text{*}}$ is the partial pressure of the pure solvent $A$ at the surface of the liquid phase.

Raoult's law relates the change in vapor pressure (the partial pressure above the liquid phase) to the mol fraction of the dissolved solute (${\chi}_{B \left(l\right)}$) and the mol fraction of the vaporized solvent (${\chi}_{A \left(v\right)}$).

Given that ${\chi}_{A \left(l\right)}$ is always between $0$ and $1$, it follows that ${P}_{A} \le {P}_{A}^{\text{*}}$. For a two-component solution, ${\chi}_{A \left(l\right)} = 1 - {\chi}_{B \left(l\right)}$, where $B$ is the other component (generally the solute).

Thus, for any quantity of solute added, the vapor pressure of the solvent above the solution decreases.

FREEZING POINT DEPRESSION AND BOILING POINT ELEVATION

The chemical potential ${\mu}_{A}$ for some solvent $A$ in the solution as it relates to the mol fraction of $A$ in an ideal solution is:

${\mu}_{A} = {\mu}_{A}^{\text{*}} + R T \ln {\chi}_{A}$

where:

• ${\mu}_{A}$ is the chemical potential of the solvent in solution.
• ${\mu}_{A}^{\text{*}}$ is the chemical potential of the pure solvent.
• ${\chi}_{A}$ is the mol fraction of $A$ in the solution. ${\chi}_{A} = 1$ if no solute is dissolved.

Chemical potential is just the molar Gibbs' free energy; it tends downwards, i.e. $\Delta {\mu}_{A} < 0$ indicates a spontaneous process, or we say that ${\mu}_{A}$ runs downhill.

Given that any typical phase transition occurs at constant pressure, consider the following graph of ${\mu}_{A}$ vs. $T$:

Now, note that in the above equation, we know that ${\chi}_{A} \le 1$. It follows that $\ln {\chi}_{A} \le 0$, and that ${\mu}_{A} < {\mu}_{A}^{\text{*}}$. Therefore, $\mu$ for the solvent must decrease when solute is dissolved into the solvent.

When $\mu$ decreases for a liquid, the liquid line in the above diagram moves downwards, which shifts ${T}_{f}$ leftwards and ${T}_{b}$ rightwards.

This simultaneously explains why the solvent freezing point decreases when a solute is dissolved into a solvent, and why the solvent boiling point also increases (by less).

OSMOTIC PRESSURE ELEVATION

Imagine a U-shaped tube separated in the middle by a semi-permeable membrane.

Osmotic pressure is the pressure that needs to be applied to prevent a pure solvent on one side from flowing to the other side due to a concentration gradient.

Osmotic pressure is given by:

$\Pi = i M R T$

where $i$ is the van't Hoff factor (the effective number of particles per dissolved solute particle), $M$ is the molarity of the solution, and $R$ and $T$ are the universal gas constant and absolute temperature, respectively.

One can readily see that if $M \uparrow$, $\Pi \uparrow$ as well. Thus, higher concentration leads to higher osmotic pressure. The van't Hoff factor accentuates that, as it increases the effective concentration of the solute ($i \ge 1$).

Conceptually, when a higher concentration of solute is on one side of the semi-permeable membrane, a higher concentration gradient exists with respect to the other side; the solvent wishes to flow into the side where the concentration is higher (the side with higher concentration "thirsts" for more solvent).

Therefore, higher concentration of solute increases the osmotic pressure, since one needs to apply a greater amount of pressure to prevent the solvent from flowing to the side with higher concentration.

## Why is the anode positive in an electrolytic cell whereas in the galvanic cell it is negative and cathode is positive?

Michael
Featured 3 months ago

This is because the terms cathode and anode describe the function of the electrode and not its charge.

#### Explanation:

Before we go into the internal workings of the two types of cell it helps if we adopt a "black box" approach to these devices.

Conventional electric current is said to flow from the positive terminal to the negative one.

This convention was adopted in the 19th Century at the time of Michael Faraday. When the electron was later discovered and shown to have a negative charge you can see from the diagram that the electron flow is from negative to positive.

This, however, is the convention we are left with.

The ANODE of the device is the terminal where conventional current flows $\textsf{\textcolor{red}{\text{in}}}$ from the outside.

This means that this is the terminal where electrons $\textsf{\textcolor{red}{\text{leave}}}$ the device.

The CATHODE of the device is the terminal where conventional current flows #sf(color(red)("out")# to the outside.

This means that this is the terminal where electrons #sf(color(red)("enter")# the device.

Now lets see how this applies to an electrolytic cell and a galvanic cell.

In an electrolytic cell electrical energy is used to cause chemical change.

A good example is the electrolysis of molten lead(II) bromride:

Bromide ions are attracted to the +ve electrode and discharged:

$\textsf{2 B {r}^{-} \rightarrow B {r}_{2} + 2 e}$

You can see from the diagram that these electrons flow into the external circuit and #sf(color(red)("leave")# the cell.

By our earlier definition this makes the +ve electrode the $\textsf{\textcolor{red}{\text{anode}}}$.

Lead(II) ions are attracted to the +ve electrode and discharged:

$\textsf{P {b}^{2 +} + 2 e \rightarrow P b}$

You can see from the diagram that electrons are $\textsf{\textcolor{red}{\text{entering}}}$ the cell from the external circuit so from our earlier definition we can say that the -ve electrode is the $\textsf{\textcolor{red}{\text{cathode}}}$

In a galvanic cell chemical energy is converted into electrical energy.

A good example is The Daniel Cell:

Zinc atoms go into solution as zinc ions:

$\textsf{Z n \rightarrow Z {n}^{2 +} + 2 e}$

So zinc forms the -ve side of the cell. You can see that these electrons flow into the external circuit and $\textsf{\textcolor{red}{\text{leave}}}$ the cell.

By our earlier definition this means that the -ve zinc electrode is now the $\textsf{\textcolor{red}{a n o \mathrm{de}}}$.

These electrons flow round the external circuit and arrive at the copper electrode.

Here copper(II) ions pick up the electrons and a deposit of copper forms:

$\textsf{C {u}^{2 +} + 2 e \rightarrow C u}$

You can see from the diagram that electrons are $\textsf{\textcolor{red}{\text{entering}}}$ the cell at the copper electrode so, from our earlier definition, this makes the +ve copper electrode the $\textsf{\textcolor{red}{\text{cathode}}}$.

The salt bridge is there to provide electrical neutrality between the two 1/2 cells.

You should be aware that oxidation is the loss of electrons and reduction is the gain of electrons.

A good way of remembering cathode and anode in a cell is to realise that:

"Oxidation occurs at the anode"

and

"Reduction occurs at the cathode"

Care must be taken when physically labelling the electrodes on any electrical device.

In a rechargeable cell such as a lead / acid accumulator the cathode and anode will change identity depending on whether it is discharging or being charged.

In summary, the terms cathode and anode depend on the function of the electrode, not its structure.

## How do you balance #K+MgBr -> KBr + Mg#?

Meave60
Featured 3 months ago

Refer to the explanation.
The balanced equation is $\text{2K" + "MgBr"_2}$$\rightarrow$$\text{2KBr" + "Mg}$.

#### Explanation:

First you need to make sure the compounds have the correct formulas. $\text{MgBr}$ is not the correct formula for magnesium bromide. The correct formula is $\text{MgBr"_2}$, because the magnesium ion has a charge of ${2}^{+}$ and each bromide ion has a charge of ${1}^{-}$. So two bromide ions are needed to bond with magnesium in order to make the compound neutral.

Now, we need to balance the equation. A balanced equation has the same number of atoms of each element on both sides. We do that by changing the amount of the reactants and products as needed by adding coefficients (numbers in front) of the formulas. Each coefficient is multiplied by the subscripts for each element. We never change a formula, so never try to balance by changing subscripts. No coefficient or subscript is understood to be $1$.

#color(purple)("K" + "MgBr"_2"#$\rightarrow$#color(purple)("KBr" + "Mg"#

There are two bromide ions on the left side and one on the right. So we need to place a coefficient of $2$ in front of $\text{KBr}$

$\textcolor{p u r p \le}{\text{K" + "MgBr"_2}}$$\rightarrow$$\textcolor{m a \ge n t a}{2} \textcolor{p u r p \le}{\text{KBr" + "Mg}}$

Now we have two potassium ions on the right side, but only one on the left side. So we need to place a coefficient of $2$ in front of $\text{K}$.

$\textcolor{m a \ge n t a}{2} \textcolor{p u r p \le}{\text{K" + "MgBr"_2}}$$\rightarrow$$\textcolor{m a \ge n t a}{2} \textcolor{p u r p \le}{\text{KBr" + "Mg}}$

Now we need to verify that the equation has the same number of atoms of each element on each side.

Left:
$\text{2 K}$
$\text{1 Mg}$
$\text{2 Br}$

Right:
$\text{2 K}$
$\text{1 Mg}$
$\text{2 Br}$

So both sides have the same number of atoms of each element, and the equation is balanced.

## Chemical equation for a reaction in which #"CO"_2# is passed through lime water for a long time?

Michael
Featured 1 month ago

See below:

#### Explanation:

When $\textsf{C {O}_{2}}$ gas is bubbled through limewater you initially get a milky precipitate of insoluble calcium carbonate:

This is an acid - base reaction and is used as a qualitative test for $\textsf{C {O}_{2}}$:

#sf(CO_2(g)+Ca(OH)_(2)(aq)rarrCaCO_3(s)+H_2O(l)#

If you continue to pass $\textsf{C {O}_{2}}$ through the suspension the acidic nature of the gas causes the precipitate to dissolve rendering the solution clear again due to the formation of soluble calcium hydrogen carbonate:

Carbonic acid is formed first:

$\textsf{C {O}_{2} + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{2} C {O}_{3} \left(a q\right)}$

This then reacts with the calcium carbonate in an acid / base reaction:

$\textsf{{H}_{2} C {O}_{3} \left(a q\right) + C a C {O}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s C a {\left(H C {O}_{3}\right)}_{2} \left(a q\right)}$

This process occurs in nature. Rain water is naturally acidic due to the presence of dissolved $\textsf{C {O}_{2}}$.

Minerals such as limestone, which is a form of calcium carbonate, are subjected to chemical weathering. This results in cave formation in limestone areas.

You will notice that the reactions are reversible. As the solution naturally evaporates this can lead to a deposit of solid calcium carbonate resulting in the formation of stalagtites and stalagmites:

Water containing dissolved $\textsf{C {a}^{2 +}}$ and $\textsf{M {g}^{2 +}}$ ions is referred to as "hard water".

A downside of this is that it forms an insoluble "scum" with soap - based detergents which is wasteful.

On the plus side hard water is good for making beer.

## How would you compare diffusion with effusion?

Nathan L.
Featured 1 week ago

Here's my interpretation.

#### Explanation:

Diffusion is when a substance disperses evenly throughout a medium; i.e. a perfume dispersing its scent through a room:

Effusion is when a substance "escapes" through a (tiny) opening; i.e. helium in a balloon escaping through the balloon's rubber network:

We can infer that the rates of diffusion and effusion of a gas depend on the speed of the particles.

In the effusion image, we notice that $\text{He}$ effuses faster than ${\text{C"_2"H"_4"O}}_{2}$....but why?

Graham's law of effusion (or diffusion) explains this:

$\frac{{r}_{1}}{{r}_{2}} = \sqrt{\frac{{M}_{2}}{{M}_{1}}}$

where

• $\frac{{r}_{1}}{{r}_{2}}$ is the ratio of the rates of effusion of a gas 1 and a gas 2

• ${M}_{1}$ and ${M}_{2}$ are the molar masses of gases 1 and 2

This equation tells us that

The rate of effusion (and diffusion) of a gas is inversely proportional to the square root of its molar mass.

Therefore, the lower the mass of a gas, the higher its rate of effusion and the higher its speed.

This makes intuitive sense, as a gas with less mass should move faster than a gas with more mass.

In terms of rates and speeds, effusion and diffusion are quite similar.

The difference between the two is that diffusion is the spreading out of a substance within a dispersing medium, and effusion is when a substance escapes through a tiny pinhole, or other hole.

Another way to differentiate between diffusion and effusion is to imagine effusion as motion of particles through a hole one-dimensionally (i.e. $\pm x$), and diffusion allows for all three dimensions (i.e. $\pm x , \pm y , \pm z$).

$\uparrow$ Credit Truong-Son N. for this addition $\uparrow$

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