60
For the reaction,
$$\ce{M -> M+ + e-}$$
the heat liberated is highest for lithium owing to its high negative $E^\circ$ value so one would think that the reaction must be most vigorous.
The reason behind the more violent reactivity of potassium rather than lithium lies in kinetics and not in thermodynamics.
No doubt, maximum energy is evolved with ...
38
It is known that ammonium nitrate decompose exothermically when heated to form nitrous oxide and water. This paper1 notes that the irreversible decomposition of ammonium nitrate occurs at the temperature range of $\pu{230-260 ^\circ C}$.
$$\ce{NH4NO3 ->[t >230 ^\circ C] N2O + 2H2O}$$
They also further noted that beyond $\pu{280 ^\circ C}$, $\ce{NH4NO3}$...
35
Can I predict the products of any chemical reaction?
In theory, yes!
Every substance has characteristic reactivity behavior. Likewise pairs and sets of substances have characteristic behavior. For example, the following combinations of substances only have one likely outcome each:
$$
\ce{HCl + NaOH -> NaCl + H2O} \\[2ex]
\ce{CH3CH2CH2OH->[$1.$ (COCl)...
30
Indeed, $\ce{Zn}$ is lower than $\ce{Na}$ in activity series of metals, so the following reaction won't take place
$$\require{cancel}\ce{Zn + 2NaOH \cancel{→} Zn(OH)2 + 2Na}$$
Remember, however, that $\ce{Zn}$ is amphoteric, so it can reacts with a strong base such as $\ce{NaOH}$ as an acid forming sodium zincate
$$\ce{Zn + 2 H2O + 2 NaOH -> Na2Zn(OH)4 ...
29
Ammonium nitrate ($\ce{NH4NO3}$) is widely used in the fertilizer industry and is one of the most concentrated forms of nitrogen fertilizer (35% of $\ce{N}$). At the same time, it has also been widely used as an explosive material for detonation in mines. Because of its explosiveness, $\ce{NH4NO3}$ is associated with various hazards including fire and ...
28
Like aniline, phenol too reacts to a very less extent during Friedel-Crafts reaction.
The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid.
In fact most substituents with lone pair would give poor yield.
The two pathways involved in the reaction with phenol reduce the overall yeild:
Phenols ...
23
First off, gold does react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the other answer
The reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects).
A nice summary of some work by Jens K. ...
19
Absolutely yes.
Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxygen-containing atmosphere. In the chamber or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give a flame. The chemical reaction would actually be the same as if ...
18
The ester carbonyl carbon is a stronger nucleophile and less prone to nucleophilic attack than the carbonyl carbon in a ketone. I think you are trying to understand why the carbonyl in a ketone typically reacts faster with a nucleophile than the carbonyl in an ester. Look at the resonance structures drawn below. Both the ketone and ester have a resonance ...
17
Here is my "old school" explanation.
Below is a drawing of the reaction coordinate for nucleophilic attack at a carbonyl carbon. The energy well for the starting carbonyl compound is shown on the left. As the positively-polarized carbonyl reacts with (forms a bond with) the nucleophile we pass over a transition state and fall into a second potential well ...
17
Relativistic effects account for gold's lack of reactivity. Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic ...
17
Though the monomers in cyanoacrylate glues contain an ester, their polymerization doesn't rely on that ester group directly. The Wikipedia article for cyanoacrylates shows the polymerization more clearly than I can easily explain in words. The many hydroxyl groups in cellulose do start polymerization effectively, and the large surface area of cotton wool ...
16
Looking at the partial charges of esters and ketones, unfortunately ron's answer is only half true. For a simple model I have chosen 3-pentanone and ethyl acetate. You can see, that the carbonyl carbon in the ketone has a smaller positive charge ($q=0.6$) than in the ester ($q=0.8$), so the latter should be more prone to nucleophilic attacks.
This is not ...
16
You might want to look up some terms, such as
noble and less noble metals
reduction potential
galvanic series
Here, reaction means that
hydrogen gas is formed
the metal is dissolved
In order to form hydrogen, protons need to be reduced to hydrogen atoms which then combine to $\ce{H2}$.
$$\ce{2 H+ + 2 e- -> H2}$$
The metal serves as an electron ...
answered Sep 19 '14 at 16:21
Klaus-Dieter Warzecha
42.8k8 gold badges95 silver badges159 bronze badges
16
You should clarify whether the iodide anion is a good nucleophile in a polar protic or polar aprotic solvent.
Also let's note that nucleophilicity is a kinetic property, while acidity/basicity are thermodynamic properties. Iodide ion's lack of basicity in water reflects its conjugate acid's ($\ce{HI}$) lack of stability; it's relatively easy to ionize its ...
15
For this approach I am basically employing Frontier Molecular Orbital Theory (FMO) to predict the reactivity of carbonyl compounds towards nucleophiles.
For the purpose of this explanation I have chosen water as nucleophile. In principle we are looking at the addition of an electron rich particle to an electron poor system. In this case, water will attack ...
14
You have to think about the whole process. When a metal loses electrons to make a metal ion the following happens:
The metallic bonds holding the metal atoms together are broken.
The metal atom loses the electrons.
The resulting metal ion is hydrated.
In your analysis you are only focusing on step 2. The enthalpy and entropy of the entire process factor ...
14
To add to @user223679's answer. Phenol can react via two pathways with acyl chlorides to give either esters, via O-acylation, or hydroxyarylketones, via C-acylation.
However, phenol esters also undergo a Fries rearrangement under Friedel-Crafts conditions to produce the C-alkylated, hydroxyarylketones. This reaction is promoted by having an excess of ...
14
Normal dioxygen ($\ce{O2}$) exists as a ground state, triplet biradical. This is an example of a molecule that, thermodynamically should be quite reactive, yet is kinetically unreactive - once again, a case of kinetics vs. thermodynamics.
Because spin must be conserved in a reaction, if ground state, triplet $\ce{O2}$ were to react with something, it ...
14
It has to do with the stability of the electrons that are ejected from the bond that breaks. In B-ketoacids, the electrons can resonate onto the oxygen and form an enol/enolate intermediate, which is much more stable than forming a carbanion on a gamma or delta ketoacid (because there is nowhere for the electrons to go other than go on the carbon atom that ...
14
First, let me point out that a rate difference of 500 is really not that large. There are solvolysis reactions with relative rate differences on the order of $\mathrm{10^{10}}$ or greater (1).
The transition state for an $\mathrm{S_{N}2}$ reaction involves hypercoordinate (or hypervalent) bonding. The transition state is, more or less, a trigonal ...
14
First,
ammonium nitrate is a kind of mixture between an oxidizer - the nitrate part - and a reducer - the ammonium one. This is at the core of your question.
The direct decomposition correctly mentioned in the answers is nevertheless a process in which something gets oxidised and something gets reduced.
In ammonium nitrate basically you have all you need - ...
13
There are a number of factors that can influence the rate of an $\mathrm{S_{N}2}$ reaction. Solvent, leaving group stability, attacking group nucleophilicity, steric factors and electronic factors.
In the series of compounds you've presented, all of these parameters are held constant except for the steric and electronic factors. Considering only steric ...
12
As you can see:
Sulfur (16) is significantly smaller than Se (34). That size difference
means that there is more room between the fluorine atoms on Se than on
Sulfur. That is what steric hindrance means. Think of steric hindrance
as limited access. The smaller the atom the less access there is.
Oxygen is so small that less than 6 fluorine atoms can ...
12
Francium is radioactive. All isotopes of francium are radioactive. The most stable isotope of francium has a half life of only 22 minutes. From the Wikipedia article I linked:
Bulk francium has never been viewed. ... However preparing such a sample is impossible, since the extreme heat of decay (the half-life of its longest-lived isotope is only 22 ...
12
As @JonCuster mentions, some materials are pretty resistant to fluorine gas at room temperature.
But out of curiosity, I checked Theodore Gray's website. He's made an effort to have a "periodic table" table with actual elements. He also has a beautiful book, called (naturally) The Elements.
Almost anything placed in the path of a stream of fluorine gas ...
12
$\Delta G = \Delta H - T \Delta S$
In the case of the $\ce{N2 + O2 -> 2NO}$ , $\Delta H$ and $\Delta S$ are both positive, so the reaction is thermodynamically favorable at high temperature (such as in lightning) but not at low temperature.
If the temperature drops to room temperature after NO is formed, it is thermodynamically favorable for NO to ...
12
The description of the hydrolysis of an acetal is accurate. The trouble is that the initial protonation event happens to such a small amount that the reaction does not proceed. We can estimate where the equilibrium of that protonation lies by looking at the pKa's of the relevant acids [Evans pKa table]. In basic solution, water (pKa 16) is the strongest acid ...
12
We need some data to answer your question. As presented in this earlier answer (which is also relevant to this question), here are the relative rates for electrophilic aromatic nitration of the halobenzenes compared to benzene itself.
\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} & \text{Relative Rate} \\
\hline
\ce{Ar-H} & 1.0 \\ \hline
\ce{Ar-F} &...
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