The rate law for the reaction
1 answer:
Explanation:
Since, the given reaction is as follows.
Hence, rate law of the reaction is as follows.
R =
As it is known that rate of a reaction depends on the initial concentration of products. So here, the rate of reaction will depend on the concentration of NO and . Since, power of the concentrations of each of these is equal to 1. Therefore, order of the reaction is equal to 1 + 1 = 2.
According to the rate law, reactants involved in the rate determining step are NO and . Hence, first step of the mechanism is the rate determining step.
Also, according to the rate of reaction doubling the concentration of NO will double the rate of reaction.
The number of reactants taking part in a single step of the reaction is known as molecularity of the reaction. Therefore, molecularity of the first step of the reaction is 2.
Both the given steps are not termolecular.
You might be interested in
Answer:
strength of hypericin in mixture = 0.42 %
Explanation:
given data
each lot = 100 g
active component hypericin = 0.3%, 0.7%, and 0.25%
solution
we get here percent strength o hypericin in the mixture that is
Hypericin contribution lot 1 = × 100
Hypericin contribution lot 1 = 0.3 g
and
Hypericin contribution lot 2 = × 100
Hypericin contribution lot 2 = 0.7 g
and
Hypericin contribution lot 3 = × 100
Hypericin contribution lot 3 = 0.25 g
so
total 300 g mixture of hypericin contain = 0.3 g + 0.7 g + 0.25 g
total 300 g mixture of hypericin = 1.25 g
so here percent strength o hypericin in mixture is
strength of hypericin in mixture = × 100
strength of hypericin in mixture = 0.42 %
Answer:
Explanation:
Hello!
In this case, since the chemical reaction between copper and nitric acid is:
By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:
However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:
Best regards!