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In a second-order reaction (one that is second order in one reactant), cutting in half the concentration of that reactant will have what effect on the reaction rate?

AThe reaction rate will remain the same.
BThe reaction rate will double.
CThe reaction rate will decrease by a factor of two.
DThe reaction rate will increase by a factor of four.
EThe reaction rate will decrease by a factor of four.

Verified Answer :

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.

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  

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

\(R'=4R\)