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denis23 [38]
3 years ago
13

In a second-order reaction (one that is second order in one reactant), cutting in half the concentration of that reactant will h

ave what effect on the reaction rate?
A) The reaction rate will remain the same.
B) The reaction rate will double.
C) The reaction rate will decrease by a factor of two.
D) The reaction rate will increase by a factor of four.
E) The reaction rate will decrease by a factor of four.
Chemistry
2 answers:
tangare [24]3 years ago
6 0

Answer: D) The reaction rate will increase by a factor of four.

Explanation:

A\rightarrow products

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.

given order =2 , thus rate law is:

R=[A]^2      (1)

On cutting the concentration of  A to half

R'=[\frac{A}{2}]^2      (2)

Dividing 2 by 1  

\frac{R'}{R}=\frac{1}{4}

R'=4R

svlad2 [7]3 years ago
4 0
A second order reaction varies with the square of the concentration of the reactant. Therefore, halving the concentration will reduce the rate of reaction by a factor of 4.
The answer is E.
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