18
Active contributors today

## This is a structured question based on Kinetics. Please, explain how to handle these type of questions. Are there any easy tips or steps? And specially, please indicate how to find the missing data in the table. That's my major problem.

Truong-Son N.
Featured 3 months ago

As an overall observation, there is no data missing from the table---it's on purpose. All the information you need is in the question information.

And the best way to learn kinetics is to have an answer key and to try practice problems. It's not easy, so you'll need to put in the practice.

Some general tips:

• Make sure you know the difference between the rate law for the reaction, the rate of reaction $r \left(t\right)$, the rate constant $k$, and the rate of disappearance of reactant $- \frac{d \left[R\right]}{\mathrm{dt}}$ or rate of appearance of product $\frac{d \left[P\right]}{\mathrm{dt}}$. These can often feel similar but they're not the same.
• When determining reaction order, don't be afraid to make up numbers and seeing what happens to $r \left(t\right)$ when you keep one initial concentration constant and change the other one.

The point is to see relationships between changes in rate and changes in reactant concentration.

• Make sure you know how to write a rate law inside and out; it's often a useful equation to know how to work with, and is central to essentially any kinetics problem.

(1) WRITING A RATE LAW

The rate law for the reaction is in general written as:

$\setminus m a t h b f \left(r \left(t\right) = k {\left[X\right]}^{m} {\left[Y\right]}^{n}\right)$

$= \setminus m a t h b f \left(- \frac{1}{{\nu}_{X}} \frac{d \left[X\right]}{\mathrm{dt}} = - \frac{1}{{\nu}_{Y}} \frac{d \left[Y\right]}{\mathrm{dt}} = \frac{1}{{\nu}_{Z}} \frac{d \left[Z\right]}{\mathrm{dt}}\right)$

for the reaction

${\nu}_{X} X + {\nu}_{Y} Y \to {\nu}_{Z} Z$

where:

• $r \left(t\right)$ is the rate as a function of time.
• $\nu$ is the stoichiometric coefficient.
• $k$ is the rate constant, in units of $\frac{1}{\text{M"^(-1 + m + n)cdot"s}}$ and ${\text{1 M" = "1 mol/dm}}^{3}$.
• $\pm \frac{d \left[\text{compound}\right]}{\mathrm{dt}}$ is the rate of disappearance of reactant (negative) or appearance of product (positive).
• $m$ and $n$ are the orders of each reactant---having no relevance to the stoichiometry of the reaction at all.

For this, ${\nu}_{X} = {\nu}_{Y} = {\nu}_{Z} = 1$, and we need to find $m$ and $n$. We don't really need $k$, apparently.

(2) FINDING THE ORDERS OF EACH REACTANT

For the reaction

$\text{X"(aq) + "Y"(aq) -> "Z} \left(a q\right) ,$

we are basically looking at what happens when we change the initial concentrations of each reactant while keeping the other constant.

We're "rigging" the reaction to see what the orders (contributions) of each reactant are with respect to the rate.

The order for each reactant is found like so:

1. Look at two trials, where one reactant's initial concentration is kept the same for both trials.
2. Now look at the other reactant. What happened to its initial concentration?
3. Now inspect the rate in $\text{mol/dm"^3cdot"s}$. What happened to the rate?
4. Based on that, you should be able to find the order.

For trials 1 and 2, halving $\left[X\right]$ halved the rate. Looking back at the rate law, basically, we are saying:

${\left[X\right]}^{m} \to {\left(\frac{\left[X\right]}{2}\right)}^{m}$ with ${\left[Y\right]}^{n} \to {\left[Y\right]}^{n}$ causes $r \left(t\right) \to \frac{r \left(t\right)}{2}$.

Therefore $m = 1$; that is, the order of $X$ is $\textcolor{b l u e}{1}$.

For trials 2 and 3, doubling $\left[Y\right]$ quadrupled $r \left(t\right)$. Basically, we are saying:

${\left[Y\right]}^{n} \to {\left(2 \left[Y\right]\right)}^{n}$ with ${\left[X\right]}^{m} \to {\left[X\right]}^{m}$ causes $r \left(t\right) \to 4 r \left(t\right)$.

So, $n = 2$, because ${2}^{2} = 4$; that is, the order of $Y$ is $\textcolor{b l u e}{2}$.

(3) FINDING A NEW RATE BASED ON NEW CONCENTRATIONS

We already got the information we needed for this in part 2.

Since we know the order of $Y$ is $2$, compare to trial 3; we doubled $\left[Y\right]$, so the rate quadruples, just like going from trial 2 to 3.

Therefore, without doing much work, $\textcolor{b l u e}{r \left(t\right) = \text{0.016 mol/dm"^3cdot"s}}$.

(4) ROLE OF D

Considering trials 4 and 5, the only trials to use $D$, the rate multiplied by $5$, for both of those two concentrations of $D$, relative to trial 3.

That tells me it's a catalyst, because a catalyst changes the mechanism of the reaction in some way, making it faster.

A common way it does so is by lowering the activation energy.

But it's surprising that the rate was the same for trials 4 and 5. I guess it means...

(5) REACTION COORDINATE DIAGRAMS

For this, that is asking you to use your answer in part 4 to draw a reaction coordinate diagram in the presence and absence of $D$. Recall that it is basically a plot of energy vs. the progress of the reaction.

As I said earlier, the activation energy, ${E}_{a}$ is lower, and additionally, the mechanism is probably different.

For this diagram, ${E}_{a}$ is the activation energy without $D$, and ${E}_{a}^{\text{*}}$ is with $D$. The dip in the curve is an intermediate formed as a result of using $D$, and we don't need to know what it actually is.

I made up the curve, but it illustrates the idea: $\setminus m a t h b f \left({E}_{a}\right)$ is lowered by $D$, the catalyst, speeding up the rate of reaction.

The products are lower in energy, indicating an energetically favorable reaction. That's an assumption, which is just saying that the reaction should happen.

(6) COMPARING TRIALS 3 AND 6

As for trial 6, we know that in comparison to trial 3, the reaction conditions are pretty much the same, except no $D$ was added for trial 6. My concern is that it's not completely true.

Everything looks similar, but remember, trials 4 and 5 had to have happened right before trial 6.

So, I would expect that some catalyst $D$ was not fully removed from solution.

That should make sense because $D$ is aqueous, so it must have been soluble, and soluble catalysts are harder to remove than insoluble catalysts. You can just scoop out a solid catalyst. You can't just do that for a dissolved catalyst!

So, probably, some residual $D$ is still there, and the reaction is still faster than it would be without $D$, but not as fast as it would be with more $D$.

## What is partial pressure of oxygen in blood?

Ernest Z.
Featured 2 months ago

${P}_{\text{O₂}}$ ≈ 100 mmHg in arterial blood, but it is different in other locations.

#### Explanation:

Here's a simplified diagram if the respiratory system.

(From www.studyblue.com)

In ambient air, ${P}_{\text{O₂}}$ = 160 mmHg.

In the alveoli

${P}_{\text{O₂}}$ in the alveoli is about 104 mmHg

The partial pressure in the alveoli is less than ${P}_{\text{O₂}}$ in ambient air because of the continual diffusion of oxygen into the alveolar capillaries.

Leaving the alveolar capillaries

Oxygen diffuses from the alveoli into the alveolar capillaries. where ${P}_{\text{O₂}}$ ≈ 100 mmHg.

In the pulmonary veins

There is no gas diffusion through veins and arteries, so ${P}_{\text{O₂}}$ is about 100 mmHg.

Entering the systemic capillaries

Blood leaving pulmonary veins enters the left atrium and is pumped from the left ventricle into the systemic circulation.

It enters the systemic capillaries with ${P}_{\text{O₂}}$ at 80 - 100 mmHg.

Leaving the systemic capillaries

${P}_{\text{O₂}}$ in the body cells is less than 40 mmHg.

Because ${P}_{\text{O₂}}$ in the systemic capillaries is greater than the partial pressure in the body cells, oxygen diffuses from the blood and into the cells.

Leaving the systemic capillaries, ${P}_{\text{O₂}}$ = 40 - 50 mmHg.

Entering the alveolar capillaries

Blood leaves the systemic capillaries and returns to the right atrium via veins.

The right ventricle then pumps the blood to the alveolar capillaries, with ${P}_{\text{O₂}}$ = 20 - 40 mmHg, and the cycle starts again.

Here's an interesting animation showing the changes in ${P}_{\text{O₂}}$ and ${P}_{\text{CO₂}}$ as the blood moves through the body.

## Estimate the value of the equilibrium constant at 610 K for each of the following reactions. ΔH∘f and S∘ for BrCl(g) is 14.6 kJ/mol and 240.0 Jmol⋅K, respectively. 2NO2(g)⇌N2O4(g)?

Ernest Z.
Featured 2 months ago

I get #K = 1.93 × 10^11# for $\text{BrCl}$ and #2.03 × 10^4# for ${\text{N"_2"O}}_{4}$.

#### Explanation:

For $\text{BrCl}$

$\text{Br"_2"(g)" + "Cl"_2"(g)" ⇌ "2BrCl(g)}$
#Δ_fH^° = "14.6 kJ·mol"^"-1"#
#ΔS^° = "240.0 J·K"^"-1""mol"^"-1"#

#Δ_fG^° = Δ_fH^° - TΔS^° = "14.6 kJ·mol"^"-1" - 610 color(red)(cancel(color(black)("K"))) × "0.2400 kJ"·color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1"= "14.6 kJ·mol"^"-1" - "146.4 kJ·mol"^"-1" = "-131.8 kJ·mol"^"-1"#

#ΔG = "-"RTlnK#

#lnK = ("-"ΔG)/(RT) = ("131 800" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 610 color(red)(cancel(color(black)("K")))) = 25.99#

#K = e^25.99 = 1.93 × 10^11#

For $\text{N"_2"O"_4}$

$\textcolor{w h i t e}{m m m m m m m m m} 2 {\text{NO"_2 ⇌ "N"_2"O}}_{4}$
#Δ_fH^°"/kJ·mol"^"-1":color(white)(ll)33.10color(white)(mmm)9.08#
#S^°"/J·K"^"-1""mol"^"-1": color(white)(ll)240.04color(white)(mm)304.38#

The formula for enthalpy of reaction is

#color(blue)(|bar(ul(color(white)(a/a) Δ_rH = sumΔ_fH_text(products) - sumΔ_fH_text(reactants)color(white)(a/a)|)))" "#

# Δ_rH = "[1(9.08) - 2(33.01)] kJ" = "-56.94 kJ"#

The formula for the entropy of reaction is

#color(blue)(|bar(ul(color(white)(a/a)Δ_rS = sumS_text(products) - sumS_text(reactants)color(white)(a/a)|)))" "#

#Δ_rS = "(1×304.38 - 2×240.04) J/K" = "-175.80 J/K"#

#color(blue)(|bar(ul(color(white)(a/a)ΔG = ΔH - TΔScolor(white)(a/a)|)))" "#

#Δ_rG = "-56.94 kJ" - 610 color(red)(cancel(color(black)("K"))) × ("-0.175 80 kJ"·color(red)(cancel(color(black)("K"^"-1")))) = "-56.94 kJ" + "107.238 kJ" = "-50.30 kJ"#

The formula for $K$ is

#color(blue)(|bar(ul(color(white)(a/a)ΔG = -RTlnKcolor(white)(a/a)|)))" "#

#lnK = ("-"ΔG)/(RT) = ("50 300" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 610 color(red)(cancel(color(black)("K")))) = 9.918#

#K = e^"9.918" = 2.03 × 10^4#

## Amino Acid Charged Groups at pH 1 and 7?

Ernest Z.
Featured 1 month ago

WARNING! Long answer. (a) The C-terminus is glycine. (b) There are 4 charged groups at pH 7. (c) At pH 1 the net charge is +2. (d) The sequence using one-letter symbols is $\text{E-M-R-T-G}$.

#### Explanation:

(a) C-terminus

In a peptide, the amino acids are written from left to right with the ${\text{NH}}_{2}$ group on the left and the $\text{C=O}$ group on the right.

The left hand amino acid is called the N-terminus, and the right hand amino acid is called the C-terminus.

Thus, in $\text{Glu-Met-Arg-Thr-Gly}$, the C-terminus is Gly (glycine).

(b) Charged groups at pH 7

There are three points to remember:

• If $\text{pH = p} {K}_{a}$, the neutral and ionic forms are present in equal amounts.
• If $\text{pH < p} {K}_{a}$ (i.e. more acidic), the protonated form will predominate.
• If $\text{pH > p} {K}_{a}$ (i.e. more basic), the non-protonated form will predominate.

Both Met (methionine) and Thr (threonine) have neutral R-groups, and they are neither N-terminal or C-terminal amino acids, so we can ignore them in our calculations.

Glu (glutamic acid) has both a basic N-terminus and an acidic R-group.

The corresponding $\text{p"K_"a}$ values are

The red line marks the pH 7 division mark.

For Gly, $\text{pH > pK"_"a}$, so the non-protonated form ($\text{COO"^"-}$) will predominate.

For Arg, $\text{pH < pK"_"a}$, so the protonated form (${\text{NH}}_{3}^{+}$) will predominate.

For the R-group of Glu, $\text{pH > pK"_"a}$, so the non-protonated form ($\text{COO"^"-}$) will predominate.

However, for the N-terminal group of Glu, $\text{pH < pK"_"a}$, so the protonated form (${\text{NH}}_{3}^{+}$) will predominate.

Thus, there are four ionized groups at pH 7.

(c) Net charge at pH 1

At pH 1, all groups have $\text{pH < pK"_"a}$, so the protonated forms will predominate..

For the C-terminal group of Gly and the R-group of Glu, the protonated forms are $\text{COOH}$.

For the N-terminal group of Glu and the R-group of Arg, the protonated forms are ${\text{NH}}_{3}^{+}$.

Thus, at pH 1 the net charge is +2.

(d) Sequence using one-letter symbols

The one-letter symbols for the amino acids are

$\text{Arg = R}$
$\text{Glu = E}$
$\text{Gly = G}$
$\text{Met = M}$
$\text{Thr = T}$

$\text{Glu-Met-Arg-Thr-Gly = E-M-R-T-G}$.

## Can someone help me?

Stefan V.
Featured 1 month ago

Here's what I got.

#### Explanation:

For part (a), use the molar mass of pyridine to calculate how many moles you have in that sample

#0.068 color(red)(cancel(color(black)("g"))) * ("1 mole C"_5"H"_5"N")/(79.10 color(red)(cancel(color(black)("g")))) = 8.597 * 10^(-4)"moles C"_5"H"_5"N"#

As you know, the number of particles needed to have one mole is given by Avogadro's number

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"particles} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Your sample of pyridine will thus contain

#8.597 * 10^(-4) color(red)(cancel(color(black)("moles C"_5"H"_5"N"))) * (6.022 * 10^(23)"molec C"_5"H"_5"N")/(1color(red)(cancel(color(black)("mole C"_5"H"_5"N"))))#

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{5.2 \cdot {10}^{20} \text{molec CH"_5"H"_5"N}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

For part (b), use the same approach to calculate the number of formula units present in your sample of zinc oxide.

You will have

#5.0 color(red)(cancel(color(black)("g"))) * "1 mole ZnO"/(81.41color(red)(cancel(color(black)("g")))) = 6.142 * 10^(-2)"moles ZnO"#

and

#6.142 * 10^(-2) color(red)(cancel(color(black)("moles ZnO"))) * (6.022 * 10^(23)"f. units ZnO")/(1color(red)(cancel(color(black)("mole ZnO"))))#

$= 3.7 \cdot {10}^{22} \text{f. units ZnO}$

The ratio between the number of molecules of pyridine and the number of formula units of zinc oxide will thus be

$\frac{5.2 \cdot {10}^{20}}{3.7 \cdot {10}^{22}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{0.014} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{w h i t e}{a}$

SIDE NOTE Notice that you can also calculate this ratio by dividing the number of moles of each compound

$\left(8.597 \cdot {10}^{- 4} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles C"_5"H"_5"N"))))/(6.142 * 10^(-2)color(red)(cancel(color(black)("moles ZnO}}}}\right) = 0.014$

That is the case because a mole is simply a fixed number of molecules / formula units.

$\textcolor{w h i t e}{a}$

For part (c), you know that zinc oxide has a surface area per gram equal to ${\text{42 m"^2"g}}^{- 1}$, which basically means that you can distribute $\text{1 g}$ of zinc oxide over a surface area of ${\text{42 m}}^{2}$.

Now, calculate how many molecules of pyridine are adsorbed per gram of zinc oxide by using the fact that you know how many molecules are adsorbed by $\text{5.0 g}$ of zinc oxide

#1 color(red)(cancel(color(black)("g ZnO"))) * (5.2 * 10^(20)"molec. C"_5"H"_5"N")/(5.0 color(red)(cancel(color(black)("g ZnO"))))#

$= 1.04 \cdot {10}^{20} \text{molec C"_5"H"_5"N}$

So, $\text{1 g}$ of zinc oxide adsorbs $1.04 \cdot {10}^{20}$ molecules of pyridine, it follows that the surface area per molecule of pyridine will be

#48"m"^2 / color(red)(cancel(color(black)("1 g ZnO"))) * color(red)(cancel(color(black)("1 g ZnO")))/(1.04 * 10^(20)"molec. C"_5"H"_5"N")#

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4.6 \cdot {10}^{- 19} \text{m"^2" / molec C"_5"H"_5"N}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answers are rounded to two sig figs.

SIDE NOTE If you want, you can convert this value from square meters per molecule to something like square nanometers per molecule

#4.6 * 10^(-19) color(red)(cancel(color(black)("m"^2)))/("1 molec. C"_5"H"_5"N") * (10^9 * 10^9"nm"^2)/(1color(red)(cancel(color(black)("m"^2))))#

$= \text{0.46 nm"^2" / molec. C"_5"H"_5"N}$

## If you use the equation Zeff=Z−S and assume that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant, S, how do you calculate Zeff for the 2p electrons in both ion?

Michael
Featured 1 week ago

You can do it like this:

#### Explanation:

Your question does not use Slater's Rules. Since you want to understand these in your sub - note I will use these rules in my answer.

Consider an atom like sodium in the gaseous state:

Sodium has 11 protons in its nucleus i.e $\textsf{Z = 11}$. The outer electron in green is attracted to the nucleus by electrostatic forces which depend on the magnitude of the charges and the distance between them.

The presence of the other 10 electrons in red has the effect of "shielding" or "screening" the outer electron from the attractive force of the nucleus.

This means that it experiences an attraction for the nucleus which is less than you would expect from the 11 protons.

This is referred to as the "effective nuclear charge" or #sf(Z_(eff)#.

So we can write:

$\textsf{{Z}_{e f f} = Z - s}$

Where $\textsf{s}$ is the screening constant and is the total screening effect of all the other electrons.

In 1930 John Slater worked out a method of calculating this screening effect which are known as "Slater's Rules".

Electrons are grouped together in increasing order of $\textsf{n}$ and $\textsf{l}$ values. $\textsf{n}$ tells you the number of the energy level and $\textsf{l}$ denotes the sub - shell such as $\textsf{s , p , d}$. Because $\textsf{s}$ and $\textsf{p}$ electrons are close in energy they are grouped together:

$\textsf{\left[1 s\right] \left[2 s , 2 p\right] \left[3 s , 3 p\right] \left[3 d\right] \left[4 s , 4 p\right] \left[4 d\right] \left[4 f\right] \left[5 s , 5 p\right]}$ and so on.

Each group has its own shielding constant. This depends on the sum of these contributions:

1.

The number of electrons in the same group.

You might think that the 2 electrons in helium are not able to shield each other since they are the same distance from the nucleus using the Bohr model:

However if you look at the probability of finding an $\textsf{s}$ electron at a given distance from the nucleus you get:

Point C is referred to as "The Bohr Radius" but you can see that there is some probability of the electron being at points closer to the nucleus such as A and B.

This means that the $\textsf{s}$ electrons in the same orbital can effectively "screen" each other. The same applies to the $\textsf{s}$ and $\textsf{p}$ electrons in the same groups.

2.

The number and type of electrons in the preceding groups.

The total shielding constant is, therefore, the overall shielding effect from $\textsf{1}$ and $\textsf{2}$.

For $\textsf{1}$ the shielding constant is the sum of :

0.35 for each other electron within the same group, except for $\textsf{1 s}$ when this is 0.3

For $\textsf{2}$ if the group is of the $\textsf{\left[\text{ns,np}\right]}$ type you add an amount of 0.85 for each electron with a principal quantum number of $\textsf{\left(n - 1\right)}$ and an amount of 1.00 for each electron with a quantum number of $\textsf{\left(n - 2\right)}$ or less.

For $\textsf{d}$ and $\textsf{f}$ electrons just add 1.00 for each electron which is closer to the nucleus.

Lets see how this applies to the examples in your comment:

#sf(""_9F^(-))# is $\textsf{\left[1 {s}^{2}\right] \left[2 {s}^{2} 2 {p}^{6}\right]}$ in which the electrons have been grouped according to the rules.

So $\textsf{Z = 9}$

Within the [2s and 2p] group a single 2p electron will be shielded by 7 other electrons each contributing 0.35 and below the group there are 2 x 1s electrons each contributing 0.85.

$\therefore$$\textsf{s = \left(7 \times 0.35\right) + \left(2 \times 0.85\right) = 4.15}$

$\therefore$$\textsf{{Z}_{e f f} = 9 - 4.15 = 4.85}$

#sf(""_11Na^(+))# is also $\textsf{\left[1 {s}^{2}\right] \left[2 {s}^{2} 2 {p}^{6}\right]}$

But this time $\textsf{Z = 11}$

So

$\textsf{{Z}_{e f f} = 11 - 4.15 = 6.85}$

This shows that a 2p electron in $\textsf{N {a}^{+}}$ experiences a greater effective nuclear charge than a 2p electron in $\textsf{{F}^{-}}$.

##### Questions
• 40 seconds ago · in Bonding
• 19 minutes ago · in Electron Configuration
• 31 minutes ago · in Hydrogen Bonds
• · 39 minutes ago · in Phases of Matter
• · 45 minutes ago · in Redox Reactions
• · 46 minutes ago · in Compounds
• · 53 minutes ago · in Enthalpy
• · 56 minutes ago
• · 59 minutes ago
• · An hour ago · in Covalent Bonding
• · An hour ago · in Bonding
• · An hour ago
• An hour ago · in pH
• · 2 hours ago
• · 2 hours ago · in The Mole
• 2 hours ago · in pH
• · 2 hours ago
• · 2 hours ago · in Changes of State
• · 3 hours ago · in Molecular Orbital Theory
• · 3 hours ago · in Density
• 3 hours ago · in Solubility Graphs
• · 3 hours ago
• · 3 hours ago · in Mixtures
• · 3 hours ago
• · 4 hours ago · in Scientific Method
• · 4 hours ago
• 4 hours ago · in Atomic Mass
• · 5 hours ago · in Equilibrium Constants
• · 5 hours ago · in Elements
• 5 hours ago · in Rate of Reactions
• · 5 hours ago · in The Periodic Table