Answer: The rate law is ![rate=k[CH_3COOC_2H_5]^1[NaOH]^1](https://tex.z-dn.net/?f=rate%3Dk%5BCH_3COOC_2H_5%5D%5E1%5BNaOH%5D%5E1)
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
For the given reaction:
k= rate constant
Rate law: ![rate=k[CH_3COOC_2H_5]^x[NaOH]^y](https://tex.z-dn.net/?f=rate%3Dk%5BCH_3COOC_2H_5%5D%5Ex%5BNaOH%5D%5Ey)
For the given rate law:
y =1 = order with respect to 
n = total order = 2
2= (x+y)
2= (x+1)
x= 1
Thus order with respect to 
is 1 and rate law is :
Answer:
The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
Explanation:
Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen using Lead (IV) oxide as a catalyst.
- The catalyst surface area is directly proportional to the reaction rate
- So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.
2. Also, Removing lead (IV) oxide from the reaction mixture the reaction rate decreased because as the catalyst is removed.
3. Using 50 cm³ of hydrogen peroxide doesn't affect the rate because the concentration of the reactant doesn't change.
4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased
So, The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
Answer:are there any abc or do you have to make an answer?
Explanation:
