Chapter 11, Problem ISP

An excellent way to make highly pure nickel metal for use in specialized steel alloys is to decompose Ni(CO)₄ by heating it in a vacuum to slightly above room temperature.

Ni(CO)₄(g) → Ni(s) + 4 CO(g)

The reaction is proposed to occur in four steps, the first of which is

Ni(CO)₄(g) → Ni(CO)₃(g) + CO(g)

Kinetic studies of this first-order decomposition reaction have been carried out between 47.3 °C and 66.0 °C to give the results in the table.*

1. (a) Determine the activation energy for this reaction.
2. (b) Ni(CO)₄ is formed by the reaction of nickel metal with carbon monoxide. Suppose that 2.05 g CO is combined with 0.125 g nickel metal. Determine the maximum mass (g) of Ni(CO)₄ that can be formed.

Replacement of CO by another molecule in Ni(CO)₄ was studied in the nonaqueous solvents toluene and hexane to understand the general principles that govern the chemistry of such compounds.*

Ni(CO)₄(g) + P(CH₃)₃ → Ni(CO)₃P(CH₃)₃ + CO

A detailed study of the kinetics of the reaction led to the mechanism

$\begin{array}{l}Step1:(slow)Ni{(}_C\to Ni{(}_C+CO\\ Step2:(fast)Ni{(}_C+P{(}_C\to Ni{(}_{C}P{(}_C\end{array}$

1. (c) Which step in the mechanism is unimolecular? Which is bimolecular?
2. (d) Add the steps of the mechanism to show that the result is the balanced equation for the observed reaction.
3. (e) Is there an intermediate in this reaction? If so, what is it?
4. (f) It was found that doubling the concentration of Ni(CO)₄ increased the reaction rate by a factor of 2. Doubling the concentration of P(CH₃)₃ had no effect on the reaction rate. Based on this information, write the rate equation for the reaction.
5. (g) Does the experimental rate equation support the proposed mechanism? Why or why not?

Interpretation Introduction

Interpretation:

The activation energy of the reaction has to be determined.  The reaction is given below.

  $Ni{\left(CO\right)}_4\left(g\right)\to Ni\left(s\right)+4CO\left(g\right)$

Concept Introduction:

The Arrhenius equation is given below.

  $k=A{e}^{-E_a/RT}$

Where, $k$ is the rate constant, $A$ is the frequency factor, $E_a$ is the activation energy, $R$ is the universal gas constant and $T$ is the temperature.

By taking $\ln$ on both the sides, the equation can be written as given below.

  $\ln k=-\frac{E_a}{RT}+\ln A$

The above equation is in the form of $y=mx+c.$  The slope of the plot of $\ln k$ versus $1/T$ is $\frac{-E_a}{R}.$

Explanation of Solution

A table has to be made as shown below by using the given data.

  $\begin{array}{ccccc}T\left({}^{\circ }C\right)& T\left(K\right)& 1/T\left(K^{-1}\right)& k\left(s^{-1}\right)& \ln k\\ 47.3& 320.5& 0.003121& 0.233& -1.457\\ 50.9& 324.1& 0.003086& 0.354& -1.038\\ 55.0& 328.2& 0.003047& 0.606& -0.501\\ 60.0& 333.2& 0.003002& 1.022& 0.02176\\ 66.0& 339.2& 0.002949& 1.873& 0.6275\end{array}$

A graph can be plotted as $\ln k$ versus $1/T$ as shown below.

Figure

From the slope of the above graph, the activation energy of the reaction can be calculated.

  $\begin{array}{l}m=\frac{-E_a}{R}\\ \Rightarrow E_a=-mR=-\left(-12190K^{-1}\right)\left(0.00831kJ{K}^{-1}mo{l}^{-1}\right)=101.3kJ/mol.\end{array}$

Therefore, the activation energy of the reaction is $101.3kJ/mol.$

Interpretation Introduction

Interpretation:

The maximum mass $\left(g\right)$ of $Ni{\left(CO\right)}_4$ that can be formed when $2.05gCO$ combines with $0.125g$ nickel metal.

Explanation of Solution

The reverse reaction of the given reaction represents the formation of $Ni{\left(CO\right)}_4.$

  $Ni\left(s\right)+4CO\left(g\right)\to Ni{\left(CO\right)}_4\left(g\right)$

Four moles of $CO$ reacts with one mole of nickel to produce one mole of $Ni{\left(CO\right)}_4.$

Determination of limiting reagent:

  $\begin{array}{l}2.05gCO\times \frac{1molCO}{28.0101gCO}\times \frac{1molNi{\left(CO\right)}_4}{4molCO}=0.0183molNi{\left(CO\right)}_4\\ \\ 0.125gNi\times \frac{1molNi}{58.6934gNi}\times \frac{1molNi{\left(CO\right)}_4}{1molNi}=0.00213molNi{\left(CO\right)}_4\end{array}$

The number of $Ni{\left(CO\right)}_4$ moles produced from nickel is smaller, thus nickel is the limiting reagent.

The mass of $0.00213molNi{\left(CO\right)}_4$ can be found out as given below.

  $0.00213molNi{\left(CO\right)}_4\times \frac{170.7338gNi{\left(CO\right)}_4}{1molNi{\left(CO\right)}_4}=0.364gNi{\left(CO\right)}_4.$

Therefore, the maximum mass $\left(g\right)$ of $Ni{\left(CO\right)}_4$ that can be formed is $0.364g.$

Interpretation Introduction

Interpretation:

The unimolecular and bimolecular step have to be chosen from the given mechanism.

  $\begin{array}{l}Step1:\left(slow\right)Ni{\left(CO\right)}_4\to Ni{\left(CO\right)}_3+CO\\ \\ Step2:\left(fast\right)Ni{\left(CO\right)}_3+P{\left(C{H}_3\right)}_3\to Ni{\left(CO\right)}_{3}P{\left(C{H}_3\right)}_3\end{array}$

Explanation of Solution

A reaction is called as elementary and unimolecular reaction if the reaction involves only one reactant.  If the reaction has exactly two reactants, either two of the same reactants or one of each of different reactants, then the reaction is bimolecular and elementary.

The first step is unimolecular as it involves exactly one reactant.  The second step is bimolecular as it involves exactly two reactants.

Interpretation Introduction

Interpretation:

The steps of the given mechanism has to be added in order to show that the result is the balanced equation for the observed reaction.

  $\begin{array}{l}Step1:\left(slow\right)Ni{\left(CO\right)}_4\to Ni{\left(CO\right)}_3+CO\\ \\ Step2:\left(fast\right)Ni{\left(CO\right)}_3+P{\left(C{H}_3\right)}_3\to Ni{\left(CO\right)}_{3}P{\left(C{H}_3\right)}_3\end{array}$

Explanation of Solution

The given mechanisms can be added as given below.

  $\begin{array}{l}Step1:\left(slow\right)Ni{\left(CO\right)}_4\to \cancel{Ni{\left(CO\right)}_3}+CO\\ \\ Step2:\left(fast\right)\underset{\_}{\cancel{Ni{\left(CO\right)}_3}+P{\left(C{H}_3\right)}_3\to Ni{\left(CO\right)}_{3}P{\left(C{H}_3\right)}_3}\\ Ni{\left(CO\right)}_4+P{\left(C{H}_3\right)}_3\to Ni{\left(CO\right)}_{3}P{\left(C{H}_3\right)}_3+CO\end{array}$

The resulting equation is the same as that given for the observed reaction.

Interpretation Introduction

Interpretation:

The intermediate involved in the reaction has to be determined.

Explanation of Solution

An intermediate is a chemical substance that is generated in an early step of a reaction and then used up in a later step.  In the given mechanism, the intermediate is $Ni{\left(CO\right)}_3.$

Interpretation Introduction

Interpretation:

Doubling the concentration of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$, the reaction rate is increased by a factor $2.$  Doubling the concentration of $P{\left(C{H}_3\right)}_3$ had no effect on the reaction rate.  Based on this information, the rate equation for the reaction has to be written.

Explanation of Solution

The reaction is given below.

  $Ni{\left(CO\right)}_4+P{\left(C{H}_3\right)}_3\to Ni{\left(CO\right)}_{3}P{\left(C{H}_3\right)}_3+CO$

Suppose the rate of the above reaction is as follows,

  $Rate=k{\left[Ni{\left(CO\right)}_4\right]}^i{\left[P{\left(C{H}_3\right)}_3\right]}^j$

Where, $k$ is the rate constant and the orders for each reactant, $iandj$, are currently unknown.

Given that, doubling the concentration of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$, the reaction rate is increased by a factor $2.$ That means the change in rate is the same as the concentration change.  This suggests that the rate is proportional to the concentration of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ and the order with respect to $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ is one.

The second condition is doubling the concentration of $P{\left(C{H}_3\right)}_3$ had no effect on the reaction rate.  Hence, the order with respect to $P{\left(C{H}_3\right)}_3$ is zero.

Now, the rate equation can be written as shown below.

  $Rate=k{\left[Ni{\left(CO\right)}_4\right]}^1{\left[P{\left(C{H}_3\right)}_3\right]}^0=k{\left[Ni{\left(CO\right)}_4\right]}^1$

Interpretation Introduction

Interpretation:

The experimental rate equation support the proposed mechanism or not, has to be explained.  The mechanism is given below.

  $\begin{array}{l}Step1:\left(slow\right)Ni{\left(CO\right)}_4\to Ni{\left(CO\right)}_3+CO\\ \\ Step2:\left(fast\right)Ni{\left(CO\right)}_3+P{\left(C{H}_3\right)}_3\to Ni{\left(CO\right)}_{3}P{\left(C{H}_3\right)}_3\end{array}$

Explanation of Solution

The experimental rate equation can be written as shown below.

  $Rate=k{\left[Ni{\left(CO\right)}_4\right]}^1{\left[P{\left(C{H}_3\right)}_3\right]}^0=k{\left[Ni{\left(CO\right)}_4\right]}^1$

The slowest step in a mechanism is the rate determining step.  Thus, according to the mechanism, the rate determining step is the first step.  The rate equation can be written as follows,

  $Rate=k{\left[Ni{\left(CO\right)}_4\right]}^1$

Thus, the experimental rate equation supports the proposed mechanism.