The following data were obtained at 25° C for the reaction [1+2+2]
C2H5Br + OH- C2H5OH + Br-
Determine the rate law for the reaction from these data.
[C2H5Br] [OH-] Rate
0.150 0.200 4.8 x 10-5
0.300 0.200 9.6 x 10-5
0.450 0.200 14.4 x 10-5
0.300 0.600 28.8 x 10-5
(ii) The decomposition of benzene diazonium chloride follows first order kinetics:
C6H5N2Cl (aq) C6H5Cl (l) + N2
The rate constant for this reaction is 0.105 s-1 at 10°C. Starting with a concentration of 0.066 M of C6H5N2Cl, calculate its concentration after 15 seconds.
(iii) What will happen to the cell potential if the concentration of the anodic solution is increased in a concentration cell. Explain
If you could explain what you don't understand about these it might make it easier to answer. These are so difficult to show on the board they don't make much sense starting from scratch.
0.014 M
(i) To determine the rate law for the reaction, we need to determine the order of reaction with respect to each reactant.
Looking at the data, we can focus on the experiments where [OH-] is constant (0.200):
Experiment 1: [C2H5Br] = 0.150 M, Rate = 4.8 x 10-5
Experiment 2: [C2H5Br] = 0.300 M, Rate = 9.6 x 10-5
Experiment 3: [C2H5Br] = 0.450 M, Rate = 14.4 x 10-5
Comparing Experiment 1 and 2, we see that when [C2H5Br] doubles (from 0.150 to 0.300 M), the rate also doubles (from 4.8 x 10-5 to 9.6 x 10-5). This indicates that the reaction is first order with respect to [C2H5Br].
Comparing Experiment 2 and 3, we see that when [C2H5Br] increases by a factor of 1.5 (from 0.300 to 0.450 M), the rate increases by a factor of 1.5 (from 9.6 x 10-5 to 14.4 x 10-5). This indicates that the reaction is first order with respect to [C2H5Br].
Now, let's look at the experiments where [C2H5Br] is constant (0.300):
Experiment 2: [OH-] = 0.200 M, Rate = 9.6 x 10-5
Experiment 4: [OH-] = 0.600 M, Rate = 28.8 x 10-5
Comparing Experiment 2 and 4, we see that when [OH-] triples (from 0.200 to 0.600 M), the rate also triples (from 9.6 x 10-5 to 28.8 x 10-5). This indicates that the reaction is first order with respect to [OH-].
Therefore, the rate law for the reaction is:
Rate = k[C2H5Br][OH-]
(ii) The decomposition of benzene diazonium chloride follows first-order kinetics with a rate constant of k = 0.105 s-1 at 10°C.
Starting with a concentration of [C6H5N2Cl] = 0.066 M, we can use the first-order integrated rate equation:
ln([C6H5N2Cl]t/[C6H5N2Cl]0) = -kt
where [C6H5N2Cl]0 is the initial concentration and [C6H5N2Cl]t is the concentration at time t.
Substituting the values:
ln([C6H5N2Cl]t/0.066) = -0.105 * 15
Solving for [C6H5N2Cl]t:
[C6H5N2Cl]t = 0.066 * e^(-0.105 * 15)
(iii) If the concentration of the anodic solution is increased in a concentration cell, the cell potential will decrease. This is because increasing the concentration of the anodic solution increases the rate of oxidation at the anode, leading to a higher flow of electrons from the anode to the cathode. This higher flow of electrons results in an increased cell current and a decrease in the cell potential. In other words, the increased concentration of the anodic solution increases the rate of the oxidation half-reaction, causing a decrease in the overall cell potential.
To determine the rate law for the reaction in question (i), we need to analyze the data provided. The rate law can be expressed in the form:
Rate = k[C2H5Br]^a[OH-]^b
where k represents the rate constant, [C2H5Br] and [OH-] are the concentrations of C2H5Br and OH-, respectively, and a and b are the reaction orders with respect to C2H5Br and OH-.
To determine the values of a and b, we can compare the rates of the reaction at different concentrations. Let's start with the first two data points:
For the first data point:
[C2H5Br] = 0.150 M
[OH-] = 0.200 M
Rate = 4.8 x 10^-5
For the second data point:
[C2H5Br] = 0.300 M
[OH-] = 0.200 M
Rate = 9.6 x 10^-5
Since the concentration of OH- is constant while the concentration of C2H5Br doubles, we can see that doubling the concentration of C2H5Br leads to a doubling of the rate. This suggests that the reaction is first-order with respect to C2H5Br.
Now, let's compare the second and third data points:
For the second data point:
[C2H5Br] = 0.300 M
[OH-] = 0.200 M
Rate = 9.6 x 10^-5
For the third data point:
[C2H5Br] = 0.450 M
[OH-] = 0.200 M
Rate = 14.4 x 10^-5
In this case, we can see that increasing the concentration of C2H5Br by 50% leads to an increase in the rate by 50%. This suggests that the reaction is first-order with respect to C2H5Br.
Finally, let's compare the second and fourth data points:
For the second data point:
[C2H5Br] = 0.300 M
[OH-] = 0.200 M
Rate = 9.6 x 10^-5
For the fourth data point:
[C2H5Br] = 0.300 M
[OH-] = 0.600 M
Rate = 28.8 x 10^-5
Here, we can see that tripling the concentration of OH- leads to tripling the rate. This suggests that the reaction is first-order with respect to OH-.
Based on these observations, the rate law for the reaction is:
Rate = k[C2H5Br][OH-]
Moving on to question (ii), we are given the rate constant (k) for the decomposition of benzene diazonium chloride. We are also given the initial concentration of C6H5N2Cl and asked to calculate its concentration after a given time.
The reaction is first order, which means that the rate of the reaction follows the equation:
ln([C6H5N2Cl]t/[C6H5N2Cl]0) = -kt
where [C6H5N2Cl]t represents the concentration at time t, [C6H5N2Cl]0 represents the initial concentration, k is the rate constant, and t is time.
Substituting the given values:
ln([C6H5N2Cl]t/0.066) = -0.105(15)
Solving for [C6H5N2Cl]t:
[C6H5N2Cl]t = 0.066 × e^(-0.105 × 15)
Calculate the value using a calculator:
[C6H5N2Cl]t ≈ 0.0179 M
So, the concentration of C6H5N2Cl after 15 seconds is approximately 0.0179 M.
Moving on to question (iii), if the concentration of the anodic solution is increased in a concentration cell, the cell potential will change. To understand why, we need to consider the Nernst equation, which relates the cell potential (E) to the concentrations of reactants and products in the cell:
E = E° - (RT/nF) * ln(Q)
In this equation, E° represents the standard cell potential, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred in the balanced equation of the cell reaction, F is Faraday's constant, and Q represents the reaction quotient.
When the concentration of the anodic solution is increased, the reaction quotient (Q) will change. As a result, the term ln(Q) in the Nernst equation will change, leading to a change in the cell potential.
Specifically, if the concentration of the anodic solution is increased, the ratio of reactants to products in the cell will change, altering the value of Q and consequently affecting the cell potential. This change can be in either direction, resulting in an increase or decrease in the cell potential.