Chapter 14, Problem 14.6A

(a)

Interpretation Introduction

To determine:

The rate law for this equation.

Interpretation Introduction

• To determine the value of ‘x’ − we can compare the rate change in experiment 1 & 2 with respect to the change in the concentration of $C_2{H}_{4}B{r}_2$As the concentration of I- is same in experiment 1 & 2 so there will be no effect of I-concentration in the rate change in experiment 1 & 2.

$\frac{{[C_2{H}_{4}B{r}_2]}_2{}^x}{{[C_2{H}_{4}B{r}_2]}_1{}^x}=\frac{{[0.343]}_2{}^x}{{[0.127]}_1{}^x}={\left(2.7\right)}^x=\frac{{[Rate]}_2}{{[Rate]}_1}=\frac{1.74*{10}^{-4}}{6.45*{10}^{-5}}=2.7$

Taking logarithm both the side-

$Log{\left(2.7\right)}^x=Log(2.7)$

$xLog\left(2.7\right)=Log(2.7)$

So, x = 1So, the order of reaction in $C_2{H}_{4}B{r}_2$ is 1
Now rate law can be further written as −
Rate = $k*{\left[C_2{H}_{4}B{r}_2\right]}^1*{\left[I^{-}\right]}^y$

• To determine the value of ‘y’: Now, we can compare the rate change in experiment 1 & 3 with respect to the change in the concentration of $C_2{H}_{4}B{r}_2$and I- both.
  $\frac{{[C_2{H}_{4}B{r}_2]}_3{}^2*{\left[I-\right]}_3{}^y}{{[C_2{H}_{4}B{r}_2]}_1{}^2*{\left[I-\right]}_1{}^y}=\frac{{[0.203]}^1*{[0.125]}^y}{{[0.127]}^1*{[0.102]}^y}=1.6*{\left(1.225\right)}^y=\frac{{[Rate]}_3}{{[Rate]}_1}=\frac{1.26*{10}^{-4}}{6.45*{10}^{-5}}=1.95$ Taking logarithm both the side- $\begin{array}{l}Log\left(1.6*{1.225}^y\right)=Log(1.95)\\ Log1.6+yLog1.225=Log(1.95)\\ 0.204+y*0.08=0.29\\ y=\frac{0.8}{0.8}=1\\ \end{array}$
  
  So the order of reaction in I- is 1
  So, the final rate law is −
  Rate = $k*{\left[C_2{H}_{4}B{r}_2\right]}^1*{\left[I^{-}\right]}^1$

(b)

To determine:

To determine the value of rate constant and its units.

Explanation:

Rate = $k*{\left[C_2{H}_{4}B{r}_2\right]}^1*{\left[I^{-}\right]}^1$

$K=\frac{6.45*{10}^{-5}M/S}{0.127*0.102M^2}=0.005M^{-1}{S}^{-1}$

(c)

To determine:

To determine the rate of formation of I₃- in the given condition.

Explanation:

Initial Rate of formation of I₃- is −
Rate= $\frac{\Delta [I_3{}^{-}]}{\Delta t}=k*{[C_2{H}_{4}B{r}_2]}^1*{\left[I^{-}\right]}^1=0.005*{(0.15)}^2=1.125*{10}^{-4}M/S$

(d)

To determine:

To determine the rate of consumption of I- in the given condition.

Explanation:

Initial Rate of loss of I- is-
Rate= $-\frac{1}{3}\frac{\Delta [I^{-}]}{\Delta t}=k*{[C_2{H}_{4}B{r}_2]}^1*{\left[I^{-}\right]}^1=0.005*{(0.15)}^2=1.125*{10}^{-4}M/S$

$-\frac{\Delta [I^{-}]}{\Delta t}=3*1.125*{10}^{-4}M/S=3.37*{10}^{-4}M/S$

Conclusion:

The rate law for the given reaction can be written in terms of $C_2{H}_{4}B{r}_2$ & $I^{-}$

Rate = $k*{\left[C_2{H}_{4}B{r}_2\right]}^x*{\left[I^{-}\right]}^y$

1. Value of rate constant (k) is 0.005 and the unit is M-¹S-1
Initial Rate of formation of I₃^-$1.1125*{10}^{-4}M/S$

Initial Rate of loss of I- is- $3.37*{10}^{-4}M/S$

(b)

Interpretation Introduction

To determine:

To determine the value of rate constant and its units.

(c)

Interpretation Introduction

To determine:

To determine the rate of formation of I₃- in the given condition.

Interpretation Introduction

To determine:

To determine the rate of consumption of I- in the given condition.