Question

The rate for the oxidation of iron(II) by cerium(IV)

$\mathit{C}{\mathit{e}}^{\mathbf{4}\mathbf{+}}\left(aq\right)\mathbf{}\mathbf{+}\mathbf{}\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}\left(aq\right)\mathbf{}\mathbf{\to }\mathbf{}\mathit{C}{\mathit{e}}^{\mathbf{3}\mathbf{+}}\left(aq\right)\mathbf{}\mathbf{+}\mathbf{}\mathit{F}{\mathit{e}}^{\mathbf{3}\mathbf{+}}\left(aq\right)$

is measured at several different initial concentrations of the two reactants:

(a) Write the rate expression for this reaction.

(b) Calculate the rate constant k and give its units.

(c) Predict the initial reaction rate for a solution in which [Ce⁴⁺] is 2.6 × 1025 M and [Fe²⁺] is 1.3 × 1025 M.

Short answer

a) Rate expression is given $Rate=k\left[F{e}^{2+}\right]\left[C{e}^{4+}\right]...........\left(2\right)$

b) Rate constant is given by $K=1.2\times {10}^{18}Mo{l}^{-1}L{s}^{-1}$

C) Rate of the reaction is found to be$4.0\times {10}^{18}$

Step-by-step solution

STEP-1: Rate Expression:

Rate expression for the reaction is given as

(a) The equation for the oxidation of iron with cerium is given as following

$C{e}^{4+}\left(aq\right)+F{e}^{2+}\left(aq\right)\to C{e}^{3+}\left(aq\right)+F{e}^{3+}\left(aq\right)$

The rate expression for the oxidation of with can be given as following.

$Rate=k{\left[F{e}^{2+}\right]}^m{\left[C{e}^{4+}\right]}^n...........\left(1\right)$

In above equation m is the order with respect to iron and n is the order with respect to cerium.

STEP-2:  Order of reaction

When the concentration of cerium is tripled from $1.1\times {10}^{-5}$to $3.4\times {10}^{-5}$the reaction is tripled so it is concluded that the rate of reaction with respect to cerium is 1^st order i.e n=1 , so the rate expression can be written as

data-custom-editor="chemistry" $Rate=k{\left[F{e}^{2+}\right]}^m\left[C{e}^{4+}\right]...........\left(2\right)$

The order of reaction with respect to iron is calculated in the following steps

STEP-3 :  Substituting concentration values

Substituting the Fe ²⁺and Ce⁺⁴concentrations in the equation 2

$\begin{array}{c}Rate1=k{\left(1.8\times {10}^{-5}\right)}^m\left(1.1\times {10}^{-5}\right)..\\ 2.0\times {10}^{-7}=k{\left(1.8\times {10}^{-5}\right)}^m\left(1.1\times {10}^{-5}\right).........\left(3\right)\\ Rate2=k{\left(2.8\times {10}^{-5}\right)}^m\left(1.1\times {10}^{-5}\right)\\ 3.1\times {10}^{-7}=k{\left(2.8\times {10}^{-5}\right)}^m\left(1.1\times {10}^{-5}\right)..........\left(4\right)\end{array}$

Step-4:  Overall order of  reaction

Dividing the equation 4 by 3

$\begin{array}{c}\frac{3.1\times {10}^{-7}}{2.0\times {10}^{-7}}=\frac{k{\left(2.8\times {10}^{-5}\right)}^m}{k{\left(1.8\times {10}^{-5}\right)}^m}\frac{\left(1.1\times {10}^{-5}\right)}{\left(1.1\times {10}^{-5}\right)}..........\left(5\right)\\ 1.55={\left(1.5\right)}^m\\ m=1\end{array}$

The order of the reaction is 1 with respect to m and hence the rate expression for the reaction is

$Rate=k\left[F{e}^{2+}\right]\left[C{e}^{4+}\right]...........\left(2\right)$

STEP-5 : Rate constant k

b) Substituting the values in rate expression to find the value of rate constant K:

$\begin{array}{l}Rate=k\left[F{e}^{2+}\right]\left[C{e}^{4+}\right]...........\left(2\right)\\ Rate=2.0\times {10}^{-7}\\ \left[F{e}^{2+}\right]=1.8\times {10}^{-5}\\ \left[C{e}^{4+}\right]=1.1\times {10}^{-5}\\ \\ Substitutingabovevaluesinequation2\\ \\ 2.0\times {10}^{-7}=K1.8\times {10}^{-5}\times 1.1\times {10}^{-5}\\ K=\frac{2.0\times {10}^{-7}}{1.8\times {10}^{-5}\times 1.1\times {10}^{-5}}\\ K=1.01\times {10}^{3}Mo{l}^{-1}L{s}^{-1}\end{array}$

The units of rate constant K is $1.01\times {10}^{3}Mo{l}^{-1}L{s}^{-1}$.

STEP-6 : Rate of  reaction

c) To find out the rate of the reaction one has to substitute the concentration values of Fe and Ce and rate constant values in the equation 2,

$\begin{array}{l}Rate=k\left[F{e}^{2+}\right]\left[C{e}^{4+}\right]...........\left(2\right)\\ Rate=1.01\times 10{}^3\times 2.6\times {10}^{-5}\times 1.3\times {10}^{-5}\\ Rate=3.41\times {10}^{-7}\end{array}$

The rate of the reaction is found to be$3.41\times {10}^{-7}$