The rate constant is numerically the same for 3 reaction of first second and third order respectively the unit of concentration being in mole per litre which reaction should be the fastest?

1 Answer
Truong-Son N.
Feb 13, 2018

This is just saying for rate laws

#r_1(t) = (k_1" ""s"^(-1))cdot[A]#

#r_2(t) = (k_2" ""M"^(-1)cdot"s"^(-1))cdot[A]^2#

#r_3(t) = (k_3" ""M"^(-2)cdot"s"^(-1))cdot[A]^3#

if #k_1 = k_2 = k_3# (which can only be by coincidence!), then you simply have larger exponents for reactions of higher order... Thus, the reaction of higher order for the SAME concentration values has the faster rate.

#r_3(t) > r_2(t) > r_1(t)#

assuming #[A]# is the same across all the reactions, and knowing that each #k# is only equal by coincidence.

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