Chapter 12, Problem 92QRT

(a)

Interpretation Introduction

Interpretation:

The mass of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ that can be formed from $\text{2}\text{.05 g CO}$ with $\text{0}\text{.125 g}$ Nickel metal has to be calculated.

Explanation of Solution

Balanced reaction is,

  ${\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}\stackrel{}{\rightleftharpoons }{\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}$

Mole of $\text{CO}$ in $\text{2}\text{.05 g CO}$ is,

    $\text{=}\text{2}\text{.05}\text{g}\text{CO}\text{×}\frac{\text{1}\text{mol}\text{CO}}{\text{28}\text{.0101}\text{CO}}\text{=0}\text{.0183}\text{mol}$

Nickel metal in $\text{0}\text{.125 g}$ Nickel is,

  $\text{=}\text{0}\text{.125}\text{g}\text{Ni×}\frac{\text{1}\text{mol}\text{Ni}}{\text{58}\text{.6934}\text{g}\text{Ni}}\text{=}\text{0}\text{.000213}\text{mol}$

The lowest mole is limiting reagent so Nickel is the reagent.

The mass of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ that can be formed from  $\text{0}\text{.000213 mol}$ Nickel is,

  $\begin{array}{l}\text{=}\text{0}\text{.000213}{\text{Ni(CO)}}_{\text{4}}\text{×}\frac{\text{170}\text{.7338}{\text{gNi(CO)}}_{\text{4}}}{\text{1}\text{mol}{\text{Ni(CO)}}_{\text{4}}}\\ \text{=}\text{0}\text{.363}\text{g}{\text{Ni(CO)}}_{\text{4}}\end{array}$

Hence, the mass of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ that can be formed from $\text{2}\text{.05 g CO}$ with $\text{0}\text{.125 g}$ Nickel metal is $\text{0}\text{.363}\text{g}$.

(b)

Interpretation Introduction

Interpretation:

The enthalpy change of decomposition reaction of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ has to be calculated.

Concept Introduction:

Hess's Law:

The enthalpy change of given reaction is calculated by subtraction of sum of enthalpy of formation reactants from sum of enthalpy of formation reactant products.

  ${\text{ΔH}}_{\text{rex}}\text{=}{\displaystyle \sum {\text{ΔH}}_{\text{product}}}\text{-}{\displaystyle \sum {\text{ΔH}}_{\text{reactant}}}$

Explanation of Solution

Balanced reaction is,

  ${\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}\stackrel{}{\rightleftharpoons }{\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}$

The standard formation enthalpy of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ gas is $\text{-602}\text{.9 kJ/mol}$

    $\begin{array}{c}{\text{ΔH}}_{\text{rex}}{\text{=ΔH}}^{\text{o}}{\text{(Ni}}_{\text{(s)}}\text{)+4}{\text{ΔH}}^{\text{o}}{\text{(CO}}_{\text{(g)}}{\text{)-ΔH}}^{\text{o}}{\text{(Ni(CO)}}_{\text{4(g)}}\\ \text{=}\text{0}\text{(kJ/mol)+4(-110}\text{.525kJ/mol)-(-602}\text{.9kJ/mol)}\\ \text{=161}\text{.9}\text{kJ/mol}\end{array}$

The enthalpy change of decomposition reaction of $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ is $\text{161}\text{.9}\text{kJ/mol}$ and it is an endothermic reaction.

(c)

Interpretation Introduction

Interpretation:

It has to be predicted whether there is an increase or a decrease in entropy when this reaction occurs.

  ${\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}\stackrel{}{\rightleftharpoons }{\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}$

Explanation of Solution

Given reaction is,   

  $\begin{array}{l}{\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}\\ \text{1}\text{1}\text{+}\text{4}\end{array}$

The mole of product is increases, when entropy of the forward reaction is increased.

In the given reaction, 5 mole of products formed from 1 mole of reactant so entropy is increased in the given reaction.

(d-i)

Interpretation Introduction

Interpretation:

The equilibrium concentration of $\text{CO}$ in the flask has to be calculated.

Concept Introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

    $\text{aA}\rightleftharpoons \text{bB}$

    $\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

    $\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

    ${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

    ${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

    ${\text{K}}_{\text{c}}$ is the equilibrium constant.

Explanation of Solution

Given reaction is,

  ${\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}$

ICE table for at given condition with given concentrations,

    $\begin{array}{l}{\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}\\ \text{initial}\text{0}\text{.01}\text{solid}\text{0}\\ \text{change}\text{-x}\text{solid}\text{+4x}\\ \text{equilibrium}\text{0}\text{.01-x}\text{4x}\end{array}$

At equilibrium,

    $\begin{array}{c}{\text{[Ni(CO)}}_{\text{4}}\text{]}\text{=}\text{0}\text{.01-x}\\ \text{=}\text{0}\text{.00001}\\ \text{x=0}\text{.01-0}\text{.00001}\\ \text{=}\text{0}\text{.01}\\ \text{[CO]}\text{=}\text{4x}\text{=}\text{0}\text{.04}\text{M}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\text{[CO]}}^{\text{4}}}{{\text{[Ni(CO)}}_{\text{4}}\text{]}}\\ \text{=}\frac{{\text{(0}\text{.04)}}^{\text{4}}}{\text{0}\text{.00001}}\text{=}\text{0}\text{.3}\end{array}$

Hence, the equilibrium concentration of $\text{CO}$ in the flask is $\text{0}\text{.04M}$.

(d-ii)

Interpretation Introduction

Interpretation:

The value of the equilibrium constant ${\text{K}}_{\text{c}}$ for this reaction at $\text{100 °C}$ has to be calculated.

Concept Introduction:

Refer part (d-i).

Explanation of Solution

Given reaction is,

  ${\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}$

ICE table for at given condition with given concentrations,

    $\begin{array}{l}{\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}\\ \text{initial}\text{0}\text{.01}\text{solid}\text{0}\\ \text{change}\text{-x}\text{solid}\text{+4x}\\ \text{equilibrium}\text{0}\text{.01-x}\text{4x}\end{array}$

At equilibrium,

    $\begin{array}{c}{\text{[Ni(CO)}}_{\text{4}}\text{]}\text{=}\text{0}\text{.01-x}\\ \text{=}\text{0}\text{.00001}\\ \text{x=0}\text{.01-0}\text{.00001}\\ \text{=}\text{0}\text{.01}\\ \text{[CO]}\text{=}\text{4x}\text{=}\text{0}\text{.04}\text{M}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\text{[CO]}}^{\text{4}}}{{\text{[Ni(CO)}}_{\text{4}}\text{]}}\\ \text{=}\frac{{\text{(0}\text{.04)}}^{\text{4}}}{\text{0}\text{.00001}}\text{=}\text{0}\text{.3}\end{array}$

Hence, the value of equilibrium constant ${\text{K}}_{\text{c}}$ for this reaction at $\text{100 °C}$ is $\text{0}\text{.3}$.

(d-iii)

Interpretation Introduction

Interpretation:

The value of the equilibrium constant ${\text{K}}_{\text{p}}$ for this reaction at $\text{100 °C}$ has to be calculated.

Explanation of Solution

Given reaction is,

  ${\text{Ni(CO)}}_{\text{4}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{Ni}}_{}{}_{\left(\text{s}\right)}\text{+}{\text{4CO}}_{\text{(g)}}$

The relationship between ${\text{K}}_{\text{c}}$ and ${\text{K}}_{\text{p}}$ is,

    ${\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\text{(RT)}}^{\text{Δn}}$

In the given balanced reaction, mole change, temperature and ${\text{K}}_c$ are,

    $\begin{array}{l}\text{Δn=}\text{4}\text{mol}{\text{CO}}_{\text{(g)}}\text{-1}\text{mol}{\text{Ni(CO)}}_{\text{4(g)}}\text{=}\text{3}\\ \text{T}\text{=}\text{100°C}\text{+}\text{273}\text{=}\text{373}\text{K}\\ {\text{K}}_{\text{c}}\text{=0}\text{.3}\end{array}$

The value of equilibrium constant ${\text{K}}_{\text{p}}$ for this reaction at $\text{100 °C}$,

    $\begin{array}{l}{\text{K}}_{\text{p}}\text{=}\text{(0}\text{.3)}\text{×}{\left[\left(\text{0}\text{.08206}\frac{\text{L×atm}}{\text{mol×K}}\right)\text{×}\left(\text{373K}\right)\right]}^{\text{3}}\\ \text{=}{\text{9×10}}^{\text{3}}\end{array}$

Hence, the value of equilibrium constant ${\text{K}}_{\text{p}}$ for this reaction at $\text{100 °C}$ is ${\text{9×10}}^{\text{3}}$.