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Chemistry 16 Online
OpenStudy (anonymous):

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. A + B yields products Trial [A] [B] Rate 1 0.15 M 0.12 M 1.3 × 10-2 M/min 2 0.30 M 0.12 M 2.6 × 10-2 M/min 3 0.15 M 0.24 M 5.2 × 10-2 M/min

OpenStudy (anonymous):

All of the following are true for a catalyst except that it provides an alternative pathway for a reaction. increases the rate of a chemical reaction. provides a lower required activation energy. is consumed during a chemical reaction.

OpenStudy (anonymous):

Your 2nd question..3 Of above seems to be correct about catalyst..Last one is false..

OpenStudy (aaronq):

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. A + B yields products Trial [A] [B] Rate 1 0.15 M 0.12 M 1.3 × 10-2 M/min 2 0.30 M 0.12 M 2.6 × 10-2 M/min 3 0.15 M 0.24 M 5.2 × 10-2 M/min to determine the exponents in the rate law (m,j), \[rate=k[A]^m[B]^j\] you need to examine 2 situations (trials) where one of the species was held constant. For example, in trial 1 and 2: [A] = 0.15M to [A]= 0.30 M, so we conclude [A] was doubled. [B] = 0.12 M for both so it was held constant in this situation we can ignore the effect of B on the rate and thus the rate law expression simplifies to: \[rate=[A]^m\] now look at the rate, in the same situation (trial 1 and 2). The rate doubled. Trial 1: rate=1=1^m Trial 2: rate=2=2^m m=1 Since the rate doubled we can conclude that the exponent should be 1. (If the rate had quadrupled, the exponent would be 2.) Now, do the same for [B].

OpenStudy (abhisar):

@superhelp101

OpenStudy (superhelp101):

lolz a year ago! haha i don't get this :/

OpenStudy (abhisar):

O_o

OpenStudy (superhelp101):

do you?

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