Chemical principle/chemistry - Essay Example

a) Fill in the missing concentrations.
Solution: From stoichiometry of the reaction one moles of dinitrogen tetroxide decomposes into tow moles of nitrogen dioxide. Using this information, the missing concentrations are filled into the blanks in bold fonts.
b) Find the rate of reaction at 30ms.
Rate of the reaction can be taken as rate of decomposition of N2O4. The rate of reaction, when plotted against concentration of N2O4, we get a straight line (figure 1, below).
This implies that the reaction is of first order. Therefore, the rate equation can be written as
............ (1)
The integrated form of this equation is ............ (2)
The rate constant of the reaction k = 24384 s-1 (negative slope of the line)
From equation (2)
Therefore, from equation (1), rate of reaction at 30 ms will be,

2. (10 points)
Determine the rate constant and order from the concentration-time data given.
(Assume that the chemical reaction is reactants à products.)
Time (s)
[Reactant], M
0
0.0350
10
0.0223
20
0.0142
50
0.0037
70
0.0015
The rate of reaction is plotted against concentration of reactant is a straight line (figure 1, below). Therefore, this is a first order reaction and the rate constant is k = 0.0384 s-1 (negative slope of the line).
3. (5 points)
Sum the following elementary steps to determine the overall stoichiometry of the reaction.
NO à N + O
O3 + O à 2O2
O2 + N à NO2
The summation of the elementary reactions results in following overall reaction:
NO + O3 NO2 + O2
4. (6 points)
Examine each of the following proposed mechanisms to determine whether it is consistent with the experimental results, and identify intermediates, if any.
2NO2 + O3 à N2O5 + O2 Rate = k[NO2][O3]
a) 2NO2 ß> N2O4 fast, reversible
N2O4 + O3 --> N2O5 + O2 slow
This mechanism is not consistent with the experimentally observed rate equation as the rate governing step or the slow reaction is the second one, therefore, this will not result in the experimentally observed rate equation.
b) NO2 + O3 à NO3 + O2 slow
NO3 + NO2 à N2O5 fast
This mechanism will result in the experimentally observed rate equation as the slow orrate governing step is the first one, which implies that the rate of the reaction will depend on the concentration of both NO2 and O3. Therefore, the intermediate will be NO3.
5. (15 points)
In an experiment, 4.95 mol of CO2, 0.050 mol of CO, and 0.050 mol of O2 are placed in a 5.0 L reaction vessel at 1400K. Calculate the reaction quotient, Q, for
CO(g) + ½ O2(g) ß> CO2(g)
Reaction quotient Q will be
a) If K for this reaction is 1.05 x 10-5, will the mixture form more CO or more CO2?
Because K > Q, therefore, the reaction will proceed in left direction towards the equilibrium, to form more of CO.
b) If this reaction is endothermic, in which direction will the reaction occur if it takes place at 1500K? How will the equilibrium constant be affected?
An endothermic reaction is favored by increasing temperature, therefore, the reaction will move in right direction to form more of CO2. This is according to Le Chatelier’s principle. The value of equilibrium constant will be much more at 1500 K than it is at 1400 K.
c) In which direction would the reaction occur if the volume of the container was decreased?
If volume of the container is decreased, then the pressure will increase. Therefore, the system will try to restore its equilibrium by reducing total number of moles in the system. This will lead to formation of more of CO2. This is according to Le Chatelier’s principle.
d) In which direction would the reaction occur if additional oxygen was added to the vessel?
Addition of more of oxygen will also make the reaction proceed in right direction, leading to formation of more of CO2 according to le Chatelier’s principle.
6. (8 points)
A vessel originally contained 0.0076M SO2, 0.0036M O2, and no SO3. After equilibrium was reached, the SO2 concentration was 0.0032M. Find Kc for
2SO2(g) + O2(g) ß> 2SO3(g)
At t = 0, 0.0076M 0.0036M 0
At equilibrium, 0.0044M 0.0020M 0.0032M
Therefore,
7. (8 points)
Exactly 2.00 atm of carbon monoxide and 3.00 atm of hydrogen are placed in reaction vessel. Kp = 5.6
CO(g) + H2(g) ß> CH2O(g)
At t = 0, 2.00 Atm 3.00 Atm 0 Atm
At equilibrium, 2 – a Atm 3 – a Atm a Atm
a) Calculate the equilibrium concentrations of all species.
or 3.43
Because, PCH2O can not be more than 3 Atm, therefore, a = 3.43 is rejected, hence a = 1.75.
Therefore, equilibrium concentration or rather equilibrium partial pressures will be following:
PCO = 2 – 1.75 Atm = 0.25 Atm
PH2 = 3 – 1.75 Atm = 1.25 Atm
PCH2O = 1.75 Atm
b) Find Kc
Because the reaction temperature is not given, it is assumed to be room temperature or 25 oC in this calculation.
Here, Dn = 1 – 2 = -1, R is Universal gas constant = 0.082 lit-Atm/K, T = 25oC~298K
8. (8 points)
A solution is made by dissolving 15.0g of sodium hydroxide in approximately 450mL of water. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to total precisely 500mL of solution.
a) Find the pH and pOH of the final solution.
Molecular mass of NaOH = 23+16+1 gm = 40 gm
Therefore, Number of moles in 15 gm NaOH = 15/40 = 0.375
Therefore, Concentration of solution will be, [NaOH] = 0.375mol/0.5lit = 0.75M
Because, NaOH is a strong electrolyte, therefore, it dissociates completely into Na+ and OH- ions. Therefore, [OH-] = 0.75
Now, pOH = -log[OH-] = -log0.75 = 0.125
b) Why must we wait for it to return to room temperature? Explain.
This is to ensure that the solution is exactly 500 mL. If water addition was done at higher temperature, there will be some loss of water due to evaporation by the time it cools down to room temperature.
9. (6 points)
A weak acid was dissolved in water to form a 0.100 M solution. If the acid is 82 % ionized, find its Ka.
Solution: Dissociation of an acid can be schematically represented by following equation:
HA H+ + A-
At t = 0, 0.1 0 0
At equilibrium, 0.1(1 – a) 0.1a 0.1a Given, a = 82% = 0.82
10. (9 points)
The following reactions take place in aqueous solution.
Predict the products and identify the acids and bases (and their conjugate species) in the reaction of
a) NH3 + CH3COOH à NH4CH3COO NH4+ + CH3COO-
Base Acid Acid Base
b) H30+ + OH- à H2O + H2O
Acid Base Acid Base
a) HSO4- + HCOO- à SO4- + HCOOH
Acid Base Base Acid
11. (8 points)
Calculate the pH of a solution prepared by adding 10.0g of NaC6H5CO2 to 100mL of 0.10M KOH.
Solution: This is a basic buffer for which pH will be determined by the concentration of the strong base i.e. KOH here.
[OH-] = [KOH] = 0.1 = 10-1
Therefore, [H+] = 10-14/[OH-] = 10-14/10-1 = 10-13
Hence, pH = -log[H+] = -log(10-13) = -(-13) = 13
1. What are the benefits of physical activity? How does the body use energy during exercise and how does this relate to weight control and cardio-respiratory fitness?
Answer: Physical activity puts ore energy demand on the metabolic system. To cater to this demand body needs to burn more of the energy packed molecules like ADP and ATP. This ultimately leads to burning of the excess fat and thus it helps in weight reduction.
Besides, to burn more of energy rich molecules our body needs more of oxygen, therefore, breathing rate is increased. This helps in keeping cardio-respiratory fitness.
2. We are in the midst of an obesity epidemic. What are your thoughts on the use of low carbohydrate versus low fat diets for weight loss? Support your argument with information you have learned so far about carbohydrates, proteins and fats.
Answer: We are in the midst of an obesity epidemic due to our own faulty life style. Due to technological advancement we have become affluent. There is no dearth of food to enjoy. Today we are consuming much more of energy through our food than we spend by the way of physical activities. Technology has provided us with vehicles so we have practically stopped walking. We have TV sets to watch while enjoying the potato chips and cold drinks. This is a simple case of demand supply mismatch with supply being more than demand and the excess is accumulating in our body.
Low carbohydrate diets are better than low fat diets for obesity control as carbohydrate is absorbed more easily in our body than fat. Also, low carbohydrate diets help in burning of excess fat. Read More