Q16.20 P - Chemistry: Molecular Nature Of Matter And Change

Question

The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both NOBr and Br2 are reddish-brown: 

$\mathbf{2}\mathit{N}\mathit{O}\mathit{B}{\mathit{r}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathbf{2}\mathit{N}{\mathit{O}}_{\left(\mathit{g}\right)}\mathbf{+}\mathit{B}{\mathit{r}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

Use the data below to answer the following: 

(a) Determine the average rate over the entire experiment.

 (b) Determine the average rate between 2.00 and 4.00 s. 

(c) Use graphical methods to estimate the initial reaction rate.

 (d) Use graphical methods to estimate the rate at 7.00 s. 

(e) At what time does the instantaneous rate equal the average rate over the entire experiment?

Time (s)	[NOBr] (mol/L)
0.00	0.0100
2.00	0.0071
4.00	0.0055
6.00	0.0045
8.00	0.0038
10.00	0.0033

Step-by-Step Solution

Answer

1. The average rate throughout the experiment will be 0.00067 mol/Ls.
2. The average rate between the time frame 2.00s and 4.00s is 0.0008 mol/Ls.
3. According to the graphical technique, the reaction's starting rate is $7.5\times {10}^{-4}\text{mol/Ls}$.
4. At 7.00 s, the rate is $1.43\times {10}^{-4}\text{mol/Ls}$.
5. The instantaneous rate will match the average rate in 4.00 s.

Step 1: (a) Average rate

The reaction rate is defined as the change in reactant or product concentrations per unit time. The average rate is the pace of reaction over time.

$\text{Average rate=}\frac{\text{Change in concentration}}{\text{Time interval}}$

The instantaneous rate refers to the change in concentration of reactants or products at a given time.

The concentration of at various points in time is as follows:

Time (s)	[NOBr] (mol/L)
0.00	0.0100
2.00	0.0071
4.00	0.0055
6.00	0.0045
8.00	0.0038
10.00	0.0033

 

The following is the reaction:

$2NOB{r}_{\left(g\right)}\to 2N{O}_{\left(g\right)}+B{r}_{2\left(g\right)}$

In terms of reactant, the reaction rate is:

$Rate=\frac{1}{2}\frac{\Delta \left[NOBr\right]}{\Delta t}$

As a result, the change in concentration between time intervals t= 0.00 s and t= 10.00 s divided by the time interval represents the average rate across the whole experiment.

$\begin{array}{rcl}\text{Average rate}& =& -\frac{\Delta \left[NOBr\right]}{\Delta t}\\ & =& -\frac{\left(0.0033M-0.0100M\right)}{\left(10.00s-0.00s\right)}\\ & =& 0.00067M/s\\ & & \end{array}$

As a result, the average rate throughout the experiment will be 0.00067 mol/Ls.

Step 2: (b) Average rate between 2.00 and 4.00 s

The change in concentrations between time intervals divided by the time interval is the average rate between time intervals, t= 2.00 s and t= 4.00 s.

$\begin{array}{rcl}\text{Average rate}& =& -\frac{\Delta \left[NOBr\right]}{\Delta t}\\ & =& -\frac{\left(0.0055M-0.007M\right)}{\left(4.00s-2.00s\right)}\\ & =& 0.0008M/s\\ & & \end{array}$

As a result, the average rate between the time frame is 0.0008 mol/Ls.

Step 3: (c) Estimate the initial reaction rate using graphical approaches.

Plotting the values of [NOBr] versus time yields the initial rate of reaction. We determine the slope of this curve by drawing a tangent at time t=0.00 s. The initial rate of response is calculated using the acquired slope.

The graph below shows the results of graphing [NOBr] on the y axis and time, t on the x-axis:

Graph between Conc. of reactant Vs time

The following is the slope of the tangent drawn at t=0.00 s:

$\begin{array}{rcl}\frac{\Delta \left[NOBr\right]}{\Delta t}& =& \frac{\left(0.0040M-0.0100M\right)}{\left(4.00s-0.00s\right)}\\ & =& -1.5\times {10}^{-3}M/s\\ & & \end{array}$

As a result, the tangent's lope is $-1.5\times {10}^{-3}M/s$.

Following is the initial reaction rate:

$\begin{array}{rcl}\text{Initial rate}& =& -\frac{1}{2}{\left[\frac{\Delta \left[NOBr\right]}{\Delta t}\right]}_{t=0.00s}\\ & =& -\frac{1}{2}\left(-1.5\times {10}^{-3}mol/Ls\right)\\ & =& 7.5\times {10}^{-4}mol/Ls\\ & & \end{array}$

According to the graphical technique, the reaction's starting rate is

Step 4: (d) Estimate the rate at 7.00 s using graphical approaches.

The graph produced by graphing [NOBr] on the Y-axis and time, t, on the x-axis is as follows:

Graph between conc. of reactant Vs time

We determine the slope of this curve by drawing a tangent at time t= 7.00 s. The rate of response is calculated using the obtained slope.

The tangent drawn at t= 7.00 s has the following slope:

$\begin{array}{rcl}\frac{\Delta \left[NOBr\right]}{\Delta t}& =& \frac{\left(0.0030M-0.0050M\right)}{\left(11.00s-4.00s\right)}\\ & =& -2.86\times {10}^{-4}M/s\\ & =& -2.86\times {10}^{-4}mol/Ls\\ & & \end{array}$

The response rate at 7.00 s is then displayed below:

$\begin{array}{rcl}\text{Rate}& =& -\frac{1}{2}{\left[\frac{\Delta \left[NOBr\right]}{\Delta t}\right]}_{t=7.00s}\\ & =& -\frac{1}{2}\left(-2.86\times {10}^{-4}mol/Ls\right)\\ & =& 1.43\times {10}^{-4}mol/Ls\\ & & \end{array}$

So, at 7.00 s, the rate is $1.43\times {10}^{-4}mol/Ls$

Step 5: (e) When does the instantaneous rate become equal to the average rate?

Let us compute the instantaneous rate by drawing a tangent at time t=7.50 s to find the time when the instantaneous rate equals the average rate.

The graph that results from graphing [NOBr] on the y axis and time, t on the x-axis is as follows:

Graph between conc. of reactant Vs time

Then, at t= 3.00 s, the instantaneous reaction rate equals the slope of the tangent drawn at t=5.00 s.

$\begin{array}{rcl}-\frac{\Delta \left[NOBr\right]}{\Delta t}& =& -\frac{\left(0.0050M-0.0063M\right)}{\left(5.00s-3.00s\right)}\\ & =& 6.5\times {10}^{-4}M/s\\ & =& 6.5\times {10}^{-4}mol/Ls\\ & & \end{array}$

Throughout the experiment, the average response rate was $6.5\times {10}^{-4}mol/Ls$ At 4.00 s, the reaction rate is $6.5\times {10}^{-4}mol/Ls$

As a result, the instantaneous rate will match the average rate in 4.00 s.