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In the reaction \(\mathrm{A \rightarrow B}\), \(\mathrm{[A]}\) is found to be 0.585 M at 81.5 s, and 0.574 M at 83.4 s. What is the average rate of the reaction during this time interval?
Average rate of reaction for reactants is:
\(\mathrm{ \dfrac{\Delta[A]}{\Delta t} =  \dfrac{0.574\, M  0.585\, M}{83.4\, s  81.5\, s} = 0.006\, M/s}\)
The initial rate of the reaction \(\mathrm{A + B \rightarrow C + D}\) is determined for different initial conditions, with the results listed in the table.
Expt  \(\mathrm{[A]}\), M  \(\mathrm{[B]}\), M  Initial Rate, Ms^{1} 
1  0.0617  0.0443  1.1167 x 10^{4} 
2  0.0617  0.0886  4.5 x 10^{4} 
3  0.1234  0.0443  2.25 x 10^{4} 
4  0.1234  0.0886  9.00 x 10^{4} 
\(\mathrm{R_1 = 1.1167 \times 10^{4} = [0.0617]^m [0.0443]^n}\)
\(\mathrm{R_2 = 4.5 \times 10^{4} = [0.0617]^m [0.0443\times 2]^n}\)
\(\mathrm{R_3 = 2.25 \times 10^{4} = [0.0617\times2]^m [0.0443]^n}\)
\(\mathrm{R_4 = 9.00 \times 10^{4} = [0.0617\times2]^m [0.0443\times2]^n}\)
Now, we can take put rate laws together and solve –
\(\mathrm{\dfrac{R_2}{R_1} = \dfrac{4.5 \times 10^{4}}{1.1167 \times 10^{4}}}\)
\(\mathrm{= \dfrac{[0.0617]^m [0.0443\times 2]^n}{[0.0617]^m [0.0443]^n}}\)
\(\mathrm{\dfrac{R_2}{R_1} = 4 = 2^n;\, n = 2}\).
Therefore, the reaction is second order with respect to \(\mathrm{B}\). Now, take \(\mathrm{\dfrac{R_3}{R_1}}\)_{:}
\(\mathrm{\dfrac{R_3}{R_1} = \dfrac{2.25 \times 10^{4}}{1.1167 \times 10^{4}} }\)
\(\mathrm{= \dfrac{[0.0617\times2]^m [0.0443]^n}{[0.0617]^m [0.0443]^n}}\)
\(\mathrm{\dfrac{R_3}{R_1} = 2 = 2^m;\, m = 1}\)
Therefore, the reaction is first order with respect to \(\mathrm{A}\).
\(\mathrm{rate\: law = k [A][B]^2}\)
The overall reaction order is the sum of the exponents – 2 for \(\mathrm{B}\) and 1 for \(\mathrm{A}\), \(\mathrm{2+1 = 3}\). Therefore the overall reaction is a third order reaction.
\(\mathrm{rate = k [A][B]^2}\)
rearrange for \(\mathrm{k}\) –
\(\mathrm{k = \dfrac{rate}{[A][B]^2}}\)
Let’s use the data from experiment 1:
\(\mathrm{k = \dfrac{1.1167\times10^{4}}{[0.0617][0.0443]^2}}\)
\(\mathrm{k = \dfrac{0.92224}{M^2s}}\)
The following rates of reaction were obtained in three experiments with the reaction
\(\ce{2NO(g) + Cl2(g) \rightarrow 2NOCl (g)}\)
What is the rate law for this reaction?
Expt  \(\ce{[NO]}\),M  \(\ce{[Cl2]}\),M  Initial Rate, Ms^{1} 
1  0.025  0.051  4.54x10^{5} 
2  0.025  0.102  9.1x10^{5} 
3  0.05  0.051  1.816x10^{4} 
Let’s write out the respective rate laws for each experiment:
\(\mathrm{R_1 = 4.54\times10^{5} = [0.025]^m[0.051]^n}\)
\(\mathrm{R_2 = 9.1\times10^{5} = [0.025]^m[0.051\times2]^n}\)
\(\mathrm{R_3 = 1.816\times10^{4} = [0.025\times2]^m[0.051]^n}\)
Now let’s solve for \(\ce{m}\) and \(\ce{n}\), the orders of the reaction with respect to \(\mathrm{[NO]}\) and \(\mathrm{[Cl2]}\)
\(\mathrm{\dfrac{R_2}{R_1} = \dfrac{9.1\times10^{5}}{4.54\times10^{5}} = 2}\)
\(\mathrm{= \dfrac{[0.025]^m[0.051\times2]^n}{[0.025]^m[0.051]^n}}\)
so, \(\mathrm{2 = 2^n}\) ; \(\mathrm{n = 1}\).
\(\mathrm{\dfrac{R_3}{R_1} = \dfrac{1.816\times10^{4}}{4.54\times10^{5}} = 4}\)
\(\mathrm{= \dfrac{[0.025\times2]^m[0.051]^n}{[0.025]^m[0.051]^n}}\)
so, \(\mathrm{4 = 2^m}\) ; \(\mathrm{m = 2}\)
Now we know that the overall rate law \(\mathrm{= k [NO]^2[Cl_2]}\) and we can solve for \(\mathrm{k}\), the rate constant with any set of experimental data.
Let’s use experiment 1 –
\(\mathrm{4.54\times10^{5} = k [0.025]^2[0.051]}\)
rearrange and solve for \(\mathrm{k}\) –
\(\begin{align}
\mathrm k &= \dfrac{4.54\times10^{5}}{[0.025]^2[0.051]} \\
&= 1.424\, \mathrm{M^{2}s^{1}}
\end{align}\)
Now we know that the \(\mathrm{rate = 1.424\,M^{2}s^{1} [NO]^2[Cl_2]}\)
The first order reaction \(\mathrm{C \rightarrow products}\) has \(\mathrm{t_{1/2} = 150\: s}\).
\(\mathrm{t_{1/2}\: (first\: order) = \dfrac{\ln 2}{k}}\)
rearrange:
\(\mathrm{k = \dfrac{\ln 2}{t_{1/2}} = \dfrac{\ln 2}{150\, s} = 0.0046/s}\)
% unreacted is \(\mathrm{\dfrac{[A]_t}{[A]_0}}\)
using the rate law:
\(\mathrm{ln \dfrac{[A]_t}{[A]_0} =  kt}\)
rearrange:
\(\begin{align}
\mathrm{\dfrac{[A]_t}{[A]_0}} &= \mathrm{e^{kt}}\\
&= \mathrm{e^{\large{(0.0046/s \,^*\, 900s)}}}\\
&= \mathrm{0.0159 \times 100 = 1.59\%\textrm{ remains unreacted}}
\end{align}\)
\(\begin{align}
\mathrm{rate} &= \mathrm{k[C]} \\
&= \mathrm{0.0046/s * 0.40\, M} \\
&= \mathrm{0.00184\, M/s}
\end{align}\)
The reaction \(\mathrm{C \rightarrow products}\) is first order in \(\mathrm{C}\).
\(\mathrm{\dfrac{2.4\, g\, C}{8} = 0.3\, g}\) (original 2.4 g of \(\mathrm{C}\) has decreased 1/8 of original value)
Notice that \(\mathrm{\dfrac{1}{8} = \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2}}\) (this means that 3 half lives have elapsed)
\(\mathrm{3 * t_{1/2} = 40\, minutes}\), \(\mathrm{t_{1/2} = 120\, minutes}\).
\(\mathrm{t_{1/2} = \dfrac{\ln 2}{k} (first\: order\: reaction)}\)
rearrange – \(\mathrm{k = \dfrac{\ln 2}{t_{1/2}}}\)
\(\mathrm{= \dfrac{\ln 2}{120\: minutes} = \dfrac{0.005776}{minutes}}\)
We know that \(\mathrm{\ln \dfrac{[A]_t}{[A]_0} = kt}\)
\(\begin{align}
&= \mathrm{ \dfrac{0.005776}{minutes} * (1.50\: hr) * \dfrac{60\: minutes}{1\: hour}}\\
&= 0.52
\end{align}\)
Now cancel out ln –
\(\begin{align}
\mathrm{\dfrac{[A]_t}{[A]_0}}&= \mathrm{e^{kt}}\\
&= \mathrm e^{0.52}\\
&= 0.5945\\
\mathrm{[A]_t}&= [A]_0 * \mathrm{e^{kt}}\\
&= 2.4 * 0.5945 \\
&=1.43\textrm{ g of C remaining undecomposed}
\end{align}\)
In the first order reaction \(\mathrm{C \rightarrow products}\), it is found that 98% of the original amount of reactant \(\mathrm{C}\) decomposes in 138 minutes. What is the halflife, t_{1}_{/2}, of this decomposition reaction?
Remember that \(\mathrm{\dfrac{[A]_t}{[A]_0}}\)is the remaining of reactant \(\mathrm{C}\) (what’s left over/ unreacted).
If 98% of the original amount of reactant \(\mathrm{C}\) has decomposed, that means there is 2% left over (0.02)
So,
\(\mathrm{0.02 = \dfrac{[A]_t}{[A]_0} = e^{kt}}\)
\(\mathrm{\ln (0.02) = \ln e^{kt}}\)
\(\mathrm{\ln (0.02) =  kt}\)
\(\mathrm{\ln (0.02) =  k (138\: minutes)}\)
\(\mathrm{\dfrac{0.028}{minutes} = k}\)
Now, using the t_{1}_{/2} equation for a first order reaction – \(\mathrm{t_{1/2} = \dfrac{\ln2}{k}}\)
\(\begin{align}
&= \mathrm{\dfrac{\ln 2}{(0.028/minutes)}}\\
&= \mathrm{2.475\: minutes}
\end{align}\)
Acetoacetic acid, \(\ce{CH3COCH2COOH}\), a reagent used in organic synthesis, decomposes in acidic solution, producing acetone and \(\ce{CO2 (g)}\).
\(\ce{CH3COCH2COOH (aq) \rightarrow CH3COCH3 (aq) + CO2 (aq)}\)
The first order decomposition has a half life of 145 minutes.
How long will it take for a sample of acetoacetic acid to be 70% decomposed?
\(\mathrm{rate = k[A]}\) (where \(\mathrm{A}\) is the acetoacetic acid)
First find the rate constant \(\mathrm{k}\) using the t_{1}_{/2} equation – \(\mathrm{t_{1/2} = \dfrac{\ln 2}{k}}\)
\(\begin{align}
\mathrm{k} &= \mathrm{\dfrac{\ln 2}{t_{1/2}}}\\
\mathrm{k} &= \mathrm{\dfrac{\ln 2}{145\: minutes}}\\
\mathrm{k} &= \mathrm{\dfrac{0.00478}{minutes}}
\end{align}\)
\(\mathrm{\dfrac{[A]_t}{[A]_0} = 0.30}\)(there is 30% remaining: 70% decomposed)
\(\mathrm{\ln\dfrac{[A]_t}{[A]_0} = \ln (0.30) =  kt}\)
\(\mathrm{\dfrac{\ln(0.30)}{k} = t}\)
\(\mathrm{\dfrac{\ln (0.30)}{0.00478/ minutes} = t}\)
\(\mathrm{t = 252\: minutes}\)
Which of these sets of date corresponds to a (a) zeroorder reaction, (b) firstorder reaction, (c) secondorder reaction?
I 
 II 
 III 

Time (s)  \(\mathrm{[A]}\) (M)  Time (s)  \(\mathrm{[A]}\) (M)  Time (s)  \(\mathrm{[A]}\) (M) 
0  1.00  0  2.00  0  0.50 
25  0.287  25  1.75  25  0.44 
50  0.223  50  1.50  50  0.40 
75  0.174  75  1.25  75  0.36 
100  0.135  100  1.00  100  0.33 
Explanation:
There are two ways to solve this problem:
Case I – First Order
Time (s)  \(\mathrm{\ln [A]}\) (M) 
0  1 
25  1.25 
50  1.5 
75  1.75 
100  2 
Case II – Zero Order

Case III – Second Order
Time (s)  \(\mathrm{\dfrac{1}{[A]}}\) (M) 
0  2 
25  2.25 
50  2.5 
75  2.75 
100  3 
i. Case I
ii. Case II
iii. Case III
What is the halflife of the first order reaction? \(\mathrm{Rate = K[A]}\) where the \(\mathrm{k = 2.3E\textrm{3}\,s^{1}}\)
\(\mathrm{t_{1/2} =301\,s}\)
Explanation:
i. \(\mathrm{T_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}}\) for a first order reaction
\(\mathrm{t_{1/2} = \dfrac{\ln2}{2.3E\textrm{3}\, s^{1}}}\)
The halflife of a first order reaction does NOT depend on the concentration. However, the halflives of both the zeroorder and the second order reactions do depend on initial concentration. In one instance, the halflife gets longer as the initial concentration increases and in the other instance, the halflife gets shorter as the initial concentration increases. Which is which?
Explanation:
\(\mathrm{t_{1/2} = \dfrac{[A]_0}{2K}}\)
\(\mathrm{t_{1/2}\propto[A]}\) where “∝” means “proportional to”
Since halflife is directly proportional to concentration, as concentration increases to infinity, the halflife increases
\(\mathrm{t_{1/2} = \dfrac{1}{k[A]_0}}\)
\(\mathrm{t_{1/2} \propto \dfrac{1}{[A]_0}}\)
Since halflife is inversely proportional to concentration, as concentration increases to infinity, the halflife decreases to 0 as \(\mathrm{\dfrac{1}{\infty} = 0}\)
Explain why:
In the reversible reaction, \(\mathrm{A + B \leftrightarrow A + B}\), the enthalpy of the forward reaction is 25 kJ/mol and the activation energy of the forward reaction is 100 kJ/mol.
Explanation: See sketch
The graph looks like a parabola with the products higher than the reactants by 25 KJ/mol. The top of the parabola is 100 kJ/mol taller than the reactants and 75 KJ/mol taller than the products. If the enthalphy were negative then the products would be lower than the reactants by 25 KJ/mol.
The decomposition of \(\mathrm{HI (g)}\) at 700 K is followed for 400 s, yielding the following data:
time (s)  \(\mathrm{[HI]}\), M  \(\mathrm{\ln [HI]}\)  \(\mathrm{\dfrac{1}{[HI]}}\), M^{1} 
0  1  0  1 
100  2/3  0.405  1.5 
200  0.5  0.301  2.0 
300  0.4  0.916  2.5 
400  1/3  1.1  3.0 
Looking at this set of data, we can see that plotting time vs. \(\mathrm{\dfrac{1}{[HI]}}\) gives you a straight line with positive slope, hence the decomposition of \(\mathrm{HI (g)}\) is a second order reaction.
We can also calculate the slope of this equation numerically, treating change in x (time) and change in y (1/molarity):
\(\mathrm{\dfrac{1.5  1}{1000}= 0.005}\),
thus proving the equation of the line valid.
For the reaction \(\mathrm{C \rightarrow products}\), the following data were obtained:
Time (Seconds)  \(\mathrm{[C]}\), M  \(\mathrm{\dfrac{1}{[C]}}\)  \(\mathrm{\ln [C]}\) 
0  0.8150  1.23  0.205 
20  0.695  1.4388  0.364 
40  0.551  1.815  0.596 
60  0.455  2.1978  0.787 
Here, you can see that the concentration in column 2 is decreasing consistently by 0.1 increments. The graph that relates time with concentration that gives a negative slope is a zero order reaction.
We can also numerically calculate the slope to be 0.0061, by taking two points and doing change in y / change in x:
\(\mathrm{\dfrac{0.695  0.8150}{200} = 0.006}\) (negative sign indicates negative slope)
The rate constant for the reaction \(\ce{H2 (g) + I2 (g) \rightarrow 2HI (g)}\) has been determined at the following temperatures: 600 K, \(\mathrm{k = 5.5 \times 10^{4}\, M^{1}s^{1}}\); 685 K, \(\mathrm{k = 2.9 \times 10^{2}\,M^{1}s^{1}}\). Calculate the activation energy for the reaction.
Use the Arrhenius equation: \(\mathrm{\ln \dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left (\dfrac{1}{T_1}  \dfrac{1}{T_2} \right )}\)
\(\mathrm{\ln \dfrac{5.5 \times 10^{4}\, M^{1}s^{1}}{2.9 \times 10^{2}\, M^{1}s^{1}} = \dfrac{E_a}{8.3145\, J\, mol^{1}K^{1}}\left (\dfrac{1}{685}  \dfrac{1}{600}\right )}\)
Rearrange to solve for \(\mathrm{E_a}\) –
\(\mathrm{E_a = \ln \dfrac{5.5 \times 10^{4}\, M^{1}s^{1}}{2.9 \times 10^{2}\, M^{1}s^{1}}*\dfrac{8.3145\, J\, mol^{1}K^{1}}{\dfrac{1}{685K}  \dfrac{1}{600K}}}\)
\(\mathrm{E_a = 1.594 \times 10^5\, J / mol}\)
Here are two statements about catalysis, are there any modifications you would make in them?
Discuss the similarities and differences between the catalytic activity of platinum metal and of an enzyme.
One similarity between the catalytic activity of platinum metal and an enzyme is that they both have metal centers that act as the active site. One difference is when they are placed in solution – the platinum metal forms a heterogeneous solution (does not dissolve) and the enzyme forms a homogeneous solution (soluble). A significant difference is that the platinum metal is nonspecific, and it can catalyze many different reactions. An enzyme is very specific and usually only works for one reaction.
The following graph depicts the effect of enzyme concentration on the rate of an enzymemeditated reaction. What are the reaction conditions necessary to account for this graph?
There must be an unlimited amount of substrate such that the enzyme is always saturated and is working at its maximum rate. If there was a finite amount of substrate, then the reaction rate will drop off as substrate is used up.
Consider the following hypothetical reaction: \(\mathrm{2A + 2B \rightarrow C + 2D}\). This reaction is second order in \(\mathrm{[A]}\) and first order in \(\mathrm{[B]}\). A threestep mechanism has been proposed and the elementary steps are as follows:
1) \(\mathrm{2A \leftrightarrow I_1 \quad (fast)}\)
2) \(\mathrm{???\quad(slow)}\)
3) \(\mathrm{I_2+ B \rightarrow C + D \quad(fast)}\)
Find the entire threestep mechanism and show that it conforms to the experimentally determined reaction order.
Note: That \(\mathrm{I_n}\)are intermediates and that the forward reaction of step 1 is in equilibrium with the reverse reaction of step 1. In other words, \(\mathrm{k_1=k_{1}}\).
\(\mathrm{Rate = k[A]^x[B]^y}\)
Since the reaction is second order in \(\mathrm{[A]}\), \(\mathrm{x = 2}\).
Since the reaction is first order in \(\mathrm{[B]}\), \(\mathrm{y = 1}\)
Therefore, the experimentally determined reaction order is as follows:
\(\mathrm{Rate = k[A]^2[B]^1}\)
We know that the three elementary steps must sum up to the overall reaction. In addition, the intermediates cannot show up in the final reaction. Solve by inspection.
When you add up the reactions, elements will cancel out and you will be left with the overall reaction.
The overall reaction when you cancel out the intermediates is:
\(\mathrm{2A + 2B \rightarrow C + 2D}\)
We know that the rate of the threestep mechanism is the reaction order of the slowest (ratedetermining) step. Note that in an elementary reaction, the coefficients are also the exponents of the reaction.
For step 2, the rate is as follows:
\(\mathrm{Rate = k[I_2][B]}\)
However, we cannot have intermediates in the rate law. So we have to solve for \(\mathrm{[I_2]}\) using step 1.
From step 1, the rate is as follows:
\(\mathrm{Rate = k_1[A]^2}\)
However, since this step is at equilibrium, the rate can also be expressed as follows:
\(\mathrm{Rate = k_{1}[I_1]}\)
Therefore,
\(\mathrm{k_1[A]^2 = k_{1}[I_1]}\)
Solving for \(\mathrm{[I_1]}\) we get
\(\mathrm{[I_1] = k_1k_{1}[A]^2}\)
We can combine \(\mathrm{k_1}\) and \(\mathrm{k_{1}}\)since they are both constants into \(\mathrm{K}\).
Thus, we get
\(\mathrm{[I_1] = K [A]^2}\)
Substituting this back into the rate in step 2 and combining \(\mathrm{k}\) and \(\mathrm{K}\) into \(\mathrm{K}\) we get
\(\mathrm{Rate = K [A]^2[B]}\)
This is the same as the experimentally determined reaction order! Success!!!
The unit of \(\mathrm{k}\) (rate of formation constant) depends on the overall order of a reaction. Here is a derivation of a general expression for the units of \(\mathrm{k}\) for a reaction of any overall order, based on the order of reaction (o), and the units of concentration (M), and time (s).
We know that rate has units of M/s always. Concentration is always in terms of molarity (M). We can look at several different cases and derive a pattern:
Case 1: The general rate equation of a first order; \(\mathrm{rate = k [A]}\).
If we just look at the units of this equation, you get: \(\mathrm{M/s = \{\textrm{units of k}\} * M}\)
rearrange for\(\mathrm{\{\textrm{units of k}\} = \dfrac{M/s}{M}= s^{1}}\)
Case 2: The general rate equation of a second order rate; \(\mathrm{= k [A]^2}\).
In terms of units: \(\mathrm{M/s = \{\textrm{units of k}\} * M^2}\)
Rearrange: \(\mathrm{\{\textrm{units of k}\} = \dfrac{M/s}{M^2} = s^{1}M^{1}}\)
Case 3: The general rate law for a zero order reaction is \(\mathrm{rate = k}\)
In terms of units: \(\mathrm{M/s = \{\textrm{units of k}\}}\)
Now, by looking at all three cases, you can see that the general expression for the units of \(\mathrm{k}\) is:
\(\mathrm{\{\textrm{units of k}\} = \dfrac{1}{M^{o1} s}}\)
where \(\mathrm{o}\) stands for order of a reaction, M stands for concentration (molarity), and s is time (seconds)
For more information on these topics check out this link: The Rate Law
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