Question

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.

Answer 1

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}`