coke Posted March 24, 2009 Posted March 24, 2009 ok, follow my reasoning: hydroxide is more reactive then oxygen proof: calcium oxide reacts with water to form calcium hydroxide iodine is more reactive then hydroxide proof: calcium hydroxide reacts with hydroiodic acid to form calcium iodide and water oxygen is more reactive then iodine proof: hydroiodic acid is oxidized by air to form iodine and water so, wheres the flaw? hydroxide makes a better ion then oxygen, iodine makes a better ion then hydroxide, and oxygen makes a better ion then iodine...
hermanntrude Posted March 24, 2009 Posted March 24, 2009 just because calcium oxide reacts with water to form hydroxide, it doesn't really mean that it's more reactive or that hydroxide "makes a better ion". What drives a reaction to happen is more complicated than that.
CaptainPanic Posted March 24, 2009 Posted March 24, 2009 This thread (click) deals with the concept of "reactivity". At the moment I write this, it's just 15 threads down in the forum index. Bottom line: "Reactivity" is not a very useful concept. You already punched your first hole in it, and as you learn more about chemistry, you will find out that "reactivity" is a concept which is a bit useless... And as hermanntrude already said - unfortunately for people who just start in chemistry, it's all a bit more complicated than that. But don't be discouraged. Chemists also often have no idea what's really happening *hides in a corner*
Kaeroll Posted March 24, 2009 Posted March 24, 2009 Bottom line: "Reactivity" is not a very useful concept. You already punched your first hole in it, and as you learn more about chemistry, you will find out that "reactivity" is a concept which is a bit useless... Welllll.... it's useful with lots of qualifiers. "X is reactive towards Y under conditions Z", as it were. But I'm just being a pedant.
coke Posted March 24, 2009 Author Posted March 24, 2009 i have a thought about this, see if it makes sense: sodium wants to be oxidized. Na2O is a stable compound, however, each oxygen is already reduced by two sodiums. An NaO radical (just in theory) is a weaker oxidizer to the sodium then a OH radical, since O is less reduced by H then by Na. This makes OH preferable to a strong reduced like sodium or calcium. With a strong reducer, hydroxides are favorable. oxygen wants to be reduced. Two Hs, as in water, kind of reduce it, but metals generally do a better job. iron hydroxide has oxygen reduced by a H and Fe. but being a strong oxidizer, it wants to be reduced even more, by two Fe (Fe is more reactive then H). So iron hydroxide oxidizes into iron oxide. With a weak reducer, oxide is favorable. So, for ionic compounds, it seems like- if the oxidizer is stronger at redox then the reducer, things are moved around so the oxidizer is happy. If the reducer is stronger, things are moved around so the reducer is happy. By happy i mean like sodium is happier as a hydroxide then an oxide.
CaptainPanic Posted March 25, 2009 Posted March 25, 2009 Welllll.... it's useful with lots of qualifiers. "X is reactive towards Y under conditions Z", as it were. But I'm just being a pedant. Indeed. The qualifiers are absolutely necessary. As I wrote in the other thread, what you have to look at is a reaction. Unless a single component will react with itself, you will need to describe the environment in which the component is present... and the pressure, temperature, possible catalysts, other components which interact (solvents, even inhibitors). In other words: to be able to say something about reactivity, you pretty much need to describe the entire experiment...
Theophrastus Posted April 1, 2009 Posted April 1, 2009 In terms of how well, or easily two compounds react its true that reactivity is really an inexistent concept in that modifying temperature, pressure, addition of a catalyst, etc. Cand easily modify this, however, when I generally think of reactivity, I think of the metal reactivity series, which can also be thought of as an inverse of the standard reduction potentials E(insert degrees symbol here) red (reduction). This is often used to predict what occurs, in the case of a redox reaction, however, as I have learned, even this does not bear 100% accuracy. (Well, science, in nature, being a "real," and not an abstract concept (like mathematics), can never be proven. For example, let us say that I am to make the statement that all sheep are white. No matter how many white sheep I find, I can never be proven correct, while the appearance of but one black sheep, shall be enough to disprove my theory) (The example I used is a rather famous one. I would state it as a quote however, I cannot remember who to attribute this to.)
UC Posted April 1, 2009 Posted April 1, 2009 In terms of how well, or easily two compounds react its true that reactivity is really an inexistent concept in that modifying temperature, pressure, addition of a catalyst, etc. Cand easily modify this, however, when I generally think of reactivity, I think of the metal reactivity series, which can also be thought of as an inverse of the standard reduction potentials E(insert degrees symbol here) red (reduction). This is often used to predict what occurs, in the case of a redox reaction, however, as I have learned, even this does not bear 100% accuracy. (Well, science, in nature, being a "real," and not an abstract concept (like mathematics), can never be proven. For example, let us say that I am to make the statement that all sheep are white. No matter how many white sheep I find, I can never be proven correct, while the appearance of but one black sheep, shall be enough to disprove my theory) (The example I used is a rather famous one. I would state it as a quote however, I cannot remember who to attribute this to.) I would read up on equilibria and standard enthalpies of formation.
hermanntrude Posted April 2, 2009 Posted April 2, 2009 balhbalhbalh.... standard reduction potentials E(insert degrees symbol here) ...blahblahblah to get a degrees symbol, simply hold down alt and press 0176, then release alt.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now