##### Questions

##### Question type

Use these controls to find questions to answer

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(t)# , the**rate constant**#k# , and the**rate of disappearance**of reactant#-(d[R])/(dt)# or**rate of appearance**of product#(d[P])/(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(t)# 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.

**DISCLAIMER:** *LONG ANSWER!*

**(1)** **WRITING A RATE LAW**

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

#\mathbf(r(t) = k[X]^m[Y]^n)#

#= \mathbf(-1/(nu_X)(d[X])/(dt) = -1/(nu_Y)(d[Y])/(dt) = 1/(nu_Z)(d[Z])/(dt))#

for the reaction

#nu_X X + nu_Y Y -> nu_Z Z# where:

#r(t)# is therateas a function of time.#nu# is the stoichiometric coefficient.#k# is therate constant, in units of#1/("M"^(-1 + m + n)cdot"s")# and#"1 M" = "1 mol/dm"^3# .#pm(d["compound"])/(dt)# is therate of disappearanceofreactant(negative) orappearanceofproduct(positive).#m# and#n# are theordersof each reactant---havingno relevanceto the stoichiometry of the reaction at all.

For this,

**(2)** **FINDING THE ORDERS OF EACH REACTANT**

For the reaction

#"X"(aq) + "Y"(aq) -> "Z"(aq),#

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:

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

For trials 1 and 2, **halving** **halved** the rate. Looking back at the rate law, basically, we are saying:

#[X]^m -> (([X])/2)^m# with#[Y]^n -> [Y]^n# causes#r(t) -> (r(t))/2# .

Therefore

For trials 2 and 3, **doubling** **quadrupled**

#[Y]^n -> (2[Y])^n# with#[X]^m -> [X]^m# causes#r(t) -> 4r(t)# .

So,

**(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

Therefore, without doing much work,

**(4)** **ROLE OF D**

Considering trials 4 and 5, the only trials to use *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...

- some excess
#D# was not used,**OR** - the limit to the catalyst's activity was reached.

**(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

*As I said earlier, the activation energy,* *is lower, and additionally, the mechanism is probably different.*

For this diagram, *without* *with*

I made up the curve, but it illustrates the idea: **is lowered** by **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 *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 **not fully removed** from solution.

That should make sense because ** soluble**, and soluble catalysts are

So, probably, some residual *as* fast as it would be with more

Featured 2 months ago

Here's a simplified diagram if the respiratory system.

(From www.studyblue.com)

In **ambient air**,

**In the alveoli**

The partial pressure in the alveoli is less than

**Leaving the alveolar capillaries**

Oxygen diffuses from the alveoli into the alveolar capillaries. where

**In the pulmonary veins**

There is no gas diffusion through veins and arteries, so

**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

**Leaving the systemic capillaries**

Because

Leaving the systemic capillaries,

**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

Here's an interesting animation showing the changes in

Featured 2 months ago

I get

**For #"BrCl"#**

**For #"N"_2"O"_4"#**

The formula for enthalpy of reaction is

#color(blue)(|bar(ul(color(white)(a/a) Î”_rH = sumÎ”_fH_text(products) - sumÎ”_fH_text(reactants)color(white)(a/a)|)))" "#

âˆ´

The formula for the entropy of reaction is

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

âˆ´

The formula for free energy change is

#color(blue)(|bar(ul(color(white)(a/a)Î”G = Î”H - TÎ”Scolor(white)(a/a)|)))" "#

âˆ´

The formula for

#color(blue)(|bar(ul(color(white)(a/a)Î”G = -RTlnKcolor(white)(a/a)|)))" "#

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

**(a) C-terminus**

In a peptide, the amino acids are written from left to right with the

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

Thus, in

**(b) Charged groups at pH 7**

There are three points to remember:

- If
#"pH = p"K_a# , the neutral and ionic forms are present in equal amounts. - If
#"pH < p"K_a# (i.e. more acidic), the protonated form will predominate. - If
#"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

The red line marks the pH 7 division mark.

For Gly,

For Arg,

For the R-group of Glu,

However, for the N-terminal group of Glu,

Thus, there are **four** ionized groups at pH 7.

**(c) Net charge at pH 1**

At pH 1, all groups have

For the C-terminal group of Gly and the R-group of Glu, the protonated forms are

For the N-terminal group of Glu and the R-group of Arg, the protonated forms are

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

**(d) Sequence using one-letter symbols**

The one-letter symbols for the amino acids are

âˆ´

Featured 1 month ago

Here's what I got.

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**

#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"particles"color(white)(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"))))#

# = color(green)(|bar(ul(color(white)(a/a)color(black)(5.2 * 10^(20)"molec CH"_5"H"_5"N")color(white)(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 * 10^(22)"f. units ZnO"#

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

#(5.2 * 10^(20))/(3.7 * 10^(22)) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.014)color(white)(a/a)|)))#

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

#(8.597 * 10^(-4) color(red)(cancel(color(black)("moles C"_5"H"_5"N"))))/(6.142 * 10^(-2)color(red)(cancel(color(black)("moles ZnO")))) = 0.014#

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

For part **(c)**, you know that zinc oxide has a *surface area per gram* equal to

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

#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 * 10^(20)"molec C"_5"H"_5"N"#

So, **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")#

# = color(green)(|bar(ul(color(white)(a/a)color(black)(4.6 * 10^(-19)"m"^2" / molec C"_5"H"_5"N")color(white)(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))))#

#="0.46 nm"^2" / molec. C"_5"H"_5"N"#

Featured 1 week ago

You can do it like this:

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

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

So we can write:

Where

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

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

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

**2.**

The number and type of electrons in the preceding groups.

The total shielding constant is, therefore, the overall shielding effect from

For

0.35 for each other electron within the same group, except for

For

For

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

So

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.

But this time

So

This shows that a 2p electron in

Ask a question
Filters

Ã—

Use these controls to find questions to answer

Unanswered

Need double-checking

Practice problems

Conceptual questions