Chemical reactions and equations 0 explanation
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Oxygen is very reactive with Alkali metals. Alkali metals are given the name alkali because the oxides of these metals react with water to form a metal hydroxide that is basic or alkaline. Lithium produces an oxide, sodium produces a peroxide, and potassium, cesium, and rubidium produce superoxides.
Antimony:SbO, SbO
Selenium:SeO, SeO
Arsenic:AsO, AsO
Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions. Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.
There are two general conditions that can give rise to zero-order rates:
Only a small fraction of the reactant molecules are in a location or state in which they are able to react, and this fraction is continually replenished from the larger pool.
When two or more reactants are involved, the concentrations of some are much greater than those of others
This situation commonly occurs when a reaction is catalyzed by attachment to a solid surface (heterogeneous catalysis) or to an enzyme.
EXAMPLE 1: DECOMPOSITION OF NITROUS OXIDE
Nitrous oxide will decompose exothermically into nitrogen and oxygen, at a temperature of approximately 575 °C
2N2O−→−−Δ,Ni2N2(g)+O2(g)(1)
This reaction in the presence of a hot platinum wire (which acts as a catalyst) is zero-order, but it follows more conventional second order kinetics when carried out entirely in the gas phase.
2N2O⟶2N2(g)+O2(g)(2)
In this case, the N2O molecules that react are limited to those that have attached themselves to the surface of the solid catalyst. Once all of the sites on the limited surface of the catalyst have been occupied, additional gas-phase molecules must wait until the decomposition of one of the adsorbed molecules frees up a surface site.
Enzyme-catalyzed reactions in organisms begin with the attachment of the substrate to the active site on the enzyme, leading to the formation of an enzyme-substrate complex. If the number of enzyme molecules is limited in relation to substrate molecules, then the reaction may appear to be zero-order.
This is most often seen when two or more reactants are involved. Thus if the reaction
A+B→products(1)
is first-order in both reactants so that
rate=k [A][B](2)
If B is present in great excess, then the reaction will appear to be zero order in B (and first order overall). This commonly happens when B is also the solvent that the reaction occurs in.
Differential Form of the Zeroth Order Rate Law
Rate=−d[A]dt=k[A]0=k=constant(3)
where Rate is the reaction rate and k is the reaction rate coefficient. In this example, the units of k are M/s. The units can vary with other types of reactions. For zero-order reactions, the units of the rate constants are always M/s. In higher order reactions, k will have different units.
Figure 1: Rate vs. time (A) and Concentration vs. time for a zero order reaction.
Integrated Form of the Zeroth Order Rate Law
Integration of the differential rate law yields the concentration as a function of time. Start with the general rate law equations
Relationship Between Half-life and Zero-order Reactions
The half-life. t1/2 , is a timescale in which each half-life represents the reduction of the initial population to 50% of its original state. We can represent the relationship by the following equation.
[A]=12[A]o(10)
Using the integrated form of the rate law, we can develop a relationship between zero-order reactions and the half-life.
[A]=[A]o−kt(11)
Substitute
12[A]o=[A]o−kt12(12)
Solve for t1/2
t1/2=[A]o2k(13)