If the concentration of acetyl chloride is increased ten times the rate of reaction is increased ten times.
The conversion of acetyl chloride to methyl acetate is a substitution reaction. Recall that a substitution reaction is one in which a moiety in a molecule is replaced by another.
In this reaction, the CH3O- ion replaces the chloride ion. In the first step, the CH3O- ion attacks the substrate in a slow step. This creates a tetrahedral intermediate. Loss of the chloride ion yields the methyl acetate product.
The rate determining step is the formation of the tetrahedral intermediate. Since the reaction is first order in the acetyl chloride, if its concentration is increased ten times the rate of reaction is increased ten times.
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Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of = 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration .
The given equilibrium reaction is,
Initially c 0
At equilibrium
The expression of will be,
where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:
Therefore, the value of equilibrium constant for this reaction is, 1.1
Answer: The rate constant for the reaction is
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = age of sample = 559 min
a = let initial amount of the reactant =
a - x = amount left after decay process =
The rate constant for the reaction is