Chapter 5, Problem 5I.27E

(a)

Interpretation Introduction

Interpretation:

The below reaction is at equilibrium or not has to be determined.

  $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right),K_c=0.56$

Concept Introduction:

Consider a gaseous reaction of the form as given below.

  $aA\left(g\right)+bB\left(g\right)\to cC\left(g\right)+dD\left(g\right)$

The expression of the reaction quotient can be written as given below.

  $Q_c=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$

Where, $\left[C\right]$ and $\left[D\right]$ are concentrations of product C and D and $\left[A\right]$ and $\left[B\right]$ are concentrations of reactant A and B.

Direction of a reaction:

• If Q < K, the reaction has tendency to proceed toward products.
• If Q = K, the reaction has achieved the equilibrium position.
• If Q > K, the reaction has tendency to proceed toward reactants.

Where, Q is reaction quotient.

Explanation of Solution

The required reaction is given below.

  $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right),K_c=0.56$

The expression for reaction quotient for the above reaction can be written as shown below,

  $Q_c=\frac{\left[PC{l}_5\right]}{\left[PC{l}_3\right]\left[C{l}_2\right]}$

The concentrations of each substance can be calculated as given below.

  $\begin{array}{l}\left[PC{l}_5\right]=\frac{Moles}{Volume}=\frac{1.50mol}{0.500L}=3M\\ \\ \left[PC{l}_3\right]=\frac{Moles}{Volume}=\frac{3.00mol}{0.500L}=6M\\ \\ \left[C{l}_2\right]=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M\end{array}$

By plugging all data in the above equation, the value of Q can be calculated as given below.

  $\begin{array}{l}{Q}_c=\frac{\left[PC{l}_5\right]}{\left[PC{l}_3\right]\left[C{l}_2\right]}\\ \\ Q_c=\frac{3}{\left(6\right)\left(1\right)}\\ \\ Q_c=0.5\end{array}$

Therefore, the reaction quotient for $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right)$ is $0.5.$

Here, $Q\ne K.$  So the reaction is not at equilibrium.

(b)

Interpretation Introduction

Interpretation:

The direction of the reaction $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right)$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The required reaction is given below.

  $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right),K_c=0.56$

The value of Q for the reaction is $\mathrm{0.5.}$

Therefore, $Q_c<K.$  So the reaction will proceed to form products.

(c)

Interpretation Introduction

Interpretation:

The equilibrium composition of the reaction $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons PC{l}_5\left(g\right)$ has to be determined.

Explanation of Solution

The concentrations of each substance can be calculated as given below.

  $\begin{array}{l}\left[PC{l}_5\right]=\frac{Moles}{Volume}=\frac{1.50mol}{0.500L}=3M\\ \\ \left[PC{l}_3\right]=\frac{Moles}{Volume}=\frac{3.00mol}{0.500L}=6M\\ \\ \left[C{l}_2\right]=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M\end{array}$

An equilibrium table can be set up as given below.

  $\begin{array}{cccc}Composition(M)& PC{l}_3& C{l}_2& PC{l}_5\\ Initialconcentration& 6& 1& 3\\ Changeinconcentration& -x& -x& +x\\ Equilibriumconcentration& 6-x& 1-x& 3+x\end{array}$

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

  $K_c=\frac{\left(3+x\right)}{\left(6-x\right)\left(1-x\right)}$

Now, the equilibrium constant expression can be solved for x.

  $\begin{array}{l}{K}_c=\frac{\left(3+x\right)}{\left(6-x\right)\left(1-x\right)}\\ \\ 0.56=\frac{\left(3+x\right)}{\left(6-x\right)\left(1-x\right)}\\ \\ 0.56x^2-4.92x+0.36=0\\ \\ x=\frac{-\left(-4.92\right)\pm \sqrt{{\left(4.92\right)}^2-4\times \left(0.56\right)\times \left(0.36\right)}}{2\times 0.56}\\ \\ x=0.071\\ \\ \left[PC{l}_5\right]=3+x=3.07mol.L^{-1}\\ \\ \left[PC{l}_3\right]=\text{6}\text{-x}=5.93mol.L^{-1}\\ \\ \left[C{l}_2\right]=1-x=0.93mol.L^{-1}\end{array}$

Therefore, the equilibrium composition of the reaction is $\left[PC{l}_5\right]=3.07mol.L^{-1},\left[PC{l}_3\right]=5.93mol.L^{-1},and\left[C{l}_2\right]=0.93mol.L.$