Determine the rate law and rate constant for reaction detailed below?

The decomposition of ethanol (#C_2H_5OH#), catalyzed by the presence of a metal surface, was studied at 600 K. Concentration versus time data were collected for this reaction, and a plot of #[C_2H_5OH]# versus time resulted in a straight line with a slope of #-4.00xx10^-5# mol/L sec.

1 Answer
Truong-Son N. · seol
Jul 1, 2018

a. #\tt{k=4.00xx10^-5mol//L*sec}#
b. #\tt{rate_(rxn)=k}#

Explanation:

1. Concentration vs time being linear: #\sf{"zero-order"}# because this corresponds to the integrated rate law given by:

#tt([A] = -kt + [A]_0)#

and because its slope has the same units as the rate of reaction. That can only mean #tt([A]^m = 1)#, i.e. #tt(m = 0)# in the rate law.

2. Zero-order rate constant #\tt{k}# is negative slope, so:

#\tt{-overbrace(-4.00xx10^-5)^"slope" = k = 4.00xx10^-5}#

3. Considering a zero-order rate law, therefore,

#\tt{rate_(rxn)=k[C_2H_5OH]^0=k}#

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