Example 3

The initial rate of reaction for the reaction E + F → G was measured at three different initial concentrations of reactants as shown in the table.

1. Determine the rate law of the reaction.

2. Determine the rate constant.

Solution

1. Comparing trials 1 and 2, [E] is doubled, while [F] and the rate constant are held constant. This comparison will allow us to determine the order of reactant E:

[latex] frac{initial rate 2 }{initial rate 1} [/latex] = [latex] {left(frac{{{mathbf [}{mathbf E}{mathbf ] }}_{{mathbf 2}}}{{{mathbf [}{mathbf E}{mathbf ] }}_{{mathbf 1}}} ight)}^y [/latex]

[latex] frac{{ m 5.47 x }{{ m 10}}^{-5} Ms^{-1} }{{ m 2.73 x }{{ m 10}}^{-5 Ms^{-1}} } [/latex] = [latex] {left(frac{{mathbf 0}.{mathbf 200}{mathbf }{mathbf M}}{{mathbf 0}.{mathbf 100}{mathbf }{mathbf M}} ight)}^y [/latex]

2.00 = 2.00^y

y = 1

Therefore, the reaction is first order with respect to [E].

Comparing trials 1 and 3, [F] is doubled, while [E] and the rate constant are held constant. This comparison will allow us to determine the order of reactant F:

[latex] frac{initial rate 3 }{initial rate 1} [/latex] = [latex] {left(frac{{{mathbf [}{mathbf F}{mathbf ] }}_{{mathbf 3}}}{{{mathbf [}{mathbf F}{mathbf ] }}_{{mathbf 1}}} ight)}^z [/latex]

[latex] frac{{ m 2.71 x }{{ m 10}}^{-5} Ms^{-1} }{{ m 2.73 x }{{ m 10}}^{-5 Ms^{-1}} } [/latex] = [latex] {left(frac{{mathbf 0}.{mathbf 200}{mathbf }{mathbf M}}{{mathbf 0}.{mathbf 100}{mathbf }{mathbf M}} ight)}^z [/latex]

0.993 = 2.00^z

z = 0

Therefore, the reaction is zero order with respect to [F].

The rate law can now be written as:

Rate = k[E]¹

2. Using the rate law we have just determined, substitute in the initial concentration values and initial rate for any trial and solve for the rate constant:

Rate = k[E]¹

Using Trial 1: [latex] { m 2.73 x }{{ m 10}}^{-5 }M s^{-1} = kleft({ m 0.100}{ m M} ight) [/latex]

k [latex] = frac{{ m 2.73 x }{{ m 10}}^{-5 }Ms^{-1}}{left({ m 0.100}{ m M} ight)} [/latex]

k = 2.73 x 10^-4 s^-1