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What is rate law? See details for full question.?
This is for a chemistry review. I was doing history homework during this lesson, so I m kind of lost.
The rate law for a chemical reaction shows the impact ___________ of the reactants on the reaction rate.
If the rate law for the reaction of reactants D, E, and F can be written as follows: Rate = k[D]^2[E], what is the reaction order for each of the three reactants?
If the concentrations of both D and E were doubled, the reaction rate would be multiplied by a factor of ____. If the concentrations of all three reactants were doubled, the rate would be multiplied by a factor of ____. Reaction oreder can t be predicted; it mist be determined __________.
Thanks! If you can add anything more about rate law, that would be awesome.
1 Answer
- Dr WLv 74 years ago
fyi..
.. (1).. a rate law is a mathematical expression relation the concentration of reactants to
.. . ... . the rate a reaction proceeds.
.. (2).. rate laws are of the form.... rate = k * [A]^n * [B]^m
.. .. .. .where
.. .. ... .. [ ] is how we express concentration in chemistry
.. .. .. .. . k is a constant.. called a "rate constant"
.. .. . . . ..m is the "order" of the reaction with respect to reactant A
... . .. . .. n is the order of the reaction with respect to reactant B
for more info, you could read through my answers here
where I derive the rate equations.
******
your question here.
.. (1).. the concentration
.. (2).. let's rewrite this as... rate = k * [D]² * [E]¹ * [F]⁰
.. .. . ..order w.r.t. D = 2
.. .. .. .order w.r.t E = 1
.. .. .. .order w.r.t F = 0
.. . .. ..can you see where the ⁰ in [F]⁰ came from? any #⁰ = 1 right?
.. (3).. rate #1 = k * [D]² * [E]¹ * [F]⁰
.. .. .. .rate #2 = k * (2[A])² * (2[E])¹ * [F]⁰ = 8 x k * [D]² * [E]¹ * [F]⁰ = 8 * rate #1
.. ... .. rate #3 = k * (2[A])² * (2[E])¹ * (2[F])⁰ = 8 x k * [D]² * [E]¹ * (2[F])⁰ = 8 * rate #1
.. (4) experimentally