Rates review problems answers and solutions:
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1.Increase the reactants’ contact surface area by changing the magnesium into little pieces.
2.Increase the temperature of the reactants.
Also :Increase (H+ ) by boiling away some of the acid solution.
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Example of an appropriate procedure
1.Graph showing the volume of hydrogen gas formed during the reaction as a function of time
Note :The curve must not be a straight line.
2.Volume of hydrogen at the 25th second and the 75th second
At the 25th second, roughly 19 mL of hydrogen gas had been formed.
At the 75th second, roughly 39 mL of hydrogen gas had been formed.
3.Average rate of formation of hydrogen gas between the 25th and the 75th second
ra = = = = 0.40 mL/s
Note :The rate may vary by 0.04 mL/s.
Answer
The average rate of formation of hydrogen gas is 0.40 mL/s.
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Example of an appropriate process
1.Change in the concentration of HCl(aq)
1.0 mol/L - 0.60 mol/L = 0.40 mol/L
2.Duration of the reaction in seconds
3.0 min = 180 s
3.Average reaction rate in moles of HCl(aq) per litre per second
Average Rate =
Average Rate = 2.2 10-3
Answer
The average reaction rate is 2.2 10-3.
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Example of an appropriate process
1.Number of moles of H2O(l) decomposed in 1 hour
18.02 g 1 mol
4.50 g ?
0.25 mol
2.Number of moles of O2 produced in 1 hour
According to the given equation, 0.25 mol of water will produce 0.125 mol of oxygen gas.
3.Average rate of production of oxygen gas in mol/h
Average Rate =
Average Rate = 0.125 mol/h
4.Time needed to produce the required amount of oxygen gas
t =
t =
t =4.0 h
Answer
The time needed is 4.0 h (240 min or 14 400 s).
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Explanation of the factor responsible for the difference noted in the rates of reaction
Examples of answers:
Nature of reactants: Wax is a larger molecule than ethanol, therefore there are more chemical bonds to break in the reaction.
Nature of reactants: Wax is a solid, ethanol is a liquid. In general, liquids react faster than solids.
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Example of solution:
1- =
Answer:The average rate of production of H2(g) between the 2nd and 3rd minute is 0.04kPa/s.
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Example of an appropriate and complete procedure
a)
b)The average reaction rate is = 2.1 mL/min
Note: Accept 0.2 mL/min. No marks if the student inverts the axes.
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Example of an appropriate procedure
a)There are three steps in the forward catalyzed reaction.
b)The peak energy of RDS for the forward uncatalyzed reaction is at point #3.
The peak energy of RDS for the forward catalyzed reaction is at point #8.
c)The activation energy of the forward uncatalyzed reaction is: +500 kJ/mol.
d)The activation energy of the reversed catalyzed reaction is: approximately +425 kJ/mol.
e)The H for the reverse reaction is: + 200 kJ/mol.
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Example of an appropriate and complete answer
1. / At 20 s, AlCl3 = 0.28 mol / (moles AlCl30.02 mol)At 10 s, AlCl3 = 0.37 mol
0.37 mol 0.28 mol = 0.09 mol AlCl3
2.0.09 mol AlCl3 = 0.14 mol Cl2
3. = 0.014 Cl2(g)
Answer:The average rate of formation of Cl2(g) between 10 s and 20 s was 0.014 .
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Example of an appropriate and complete solution
In the second experiment, the marble slab is sliced into two parts. As a consequence, the surface area of the marble will be doubled and the rate will double.
Since this is experimental data, some variation ( 1 sec) in the data and in the answers must be allowed. The final time must be close to 23 sec ().
Answer:
Volume of gas / 5 mL / 10 mL / 15 mL / 20 mL / 25 mL / 30 mLElapsed time / 4 s / 8 s / 11 s / 15 s / 19 s / 22 s
The surface area of the four edges of the slab is not doubled when the slab is cut but this will be a very minor factor given that the slab of marble is said to be “very thin”.
Calculate the rate:
Average rate=
=
Answer:The average rate of reaction in the second experiment is .
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Example of an appropriate and complete answer
For 0-10 s
moles Fe(s) used = 1.10 mol 0.75 mol
= 0.35 mol
moles CO(g) present
Fe(s) : Co(g)= 3:2
moles CO(g)= 0.35 mol
= 0.553 mol
At 10 s, 0.53 moles of CO(g) is present.
Repeat calculations for subsequent intervals.
The following table summarizes the results.
Time (s) / Fe(s) present (mol) / Fe(s) used (mol) / CO(g) present (mol)0 / 1.10 / 0 / 0
10 / 0.75 / 0.35 / 0.53
20 / 0.60 / 0.50 / 0.75
30 / 0.50 / 0.60 / 0.90
40 / 0.45 / 0.65 / 0.98
50 / 0.42 / 0.68 / 1.02
Note:All extended answers have been worked out using significant figures. However, do not penalize students if they do not consider significant figures.
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Example of an appropriate and complete solution
Justification:Decreasing the temperature increases the probability of finding molecules with lower energy levels; hence, the curve shifts to the left.
When a catalyst is added, the reaction follows an alternate pathway that has a lower activation energy. This results in the activation energy (Ea) shifting to the left.
Note:All extended answers have been worked out using significant figures. However, do not penalize students if they do not consider significant figures.
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1 and 4 only
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The situation in which the beaker on the right produces the more rapid reaction is : the fourth.
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Example of an appropriate and complete procedure
a)In B, an increase in temperature increases the fraction of molecules with the
activation energy required to react.
b)In C, the activation energy is lowered by the addition of a catalyst.
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Example of an appropriate procedure
Reaction n° / Slower Reaction / Justification1 / A / The magnesium ribbon has a lower surface area than the powder. Therefore, there will be fewer effective collisions with the reacting magnesium atoms in the air and the reaction will be slower.
2 / A / When a catalyst is added, the activation energy is lowered and there is a greater number of collisions between reactant molecules with the necessary kinetic energy to form products. Therefore, there will be fewer effective collisions without a catalyst.
3 / B / The reaction will be slower at 20°C due to fewer effective collisions with sufficient kinetic energy.
4 / A / The lower the concentration of the oxygen, the fewer the collisions that will occur. Since air has a lower concentration of oxygen, there will be fewer collisions and the reaction will be slower.
5 / B / The more complex the nature of the reactants, the more bonds that have to be broken. This increases the number of effective collisions needed. This makes the reaction slower.
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B
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D
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C
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D