Student ID Number: ________________________________
Instructions: Please use pen. Calculators allowed.
Part A: (3 marks each)
Q. 1| Q.2| Q.3| Q.4| Q.5| Q.6|
Part B: (10 marks each)
Q.1| Q.2| Q.3| Q4.|
Total: / 52 marks
Part A: Short Answer. Only answer 4 of the 6 questions.
(3 marks each)
A1.Under standard state conditions, predict which has the larger standard entropy and give a reason why: 1 mole each of NO and NO2 ?
ANS: Both NO and NO2 are gases under standard conditions. Each molecule of NO2 has three atoms, and each molecule of NO has two atoms. Thus, NO2 should have the larger standard molar entropy than NO. Molecules with many atoms has more variations in how it can vibrate than a molecule with fewer atoms. This means more ways to distribute energy- or greater energy dispersal at any given temperature and larger entropy.
A2.ΔGo for reaction A is -50 kJ mol-1. ΔGo for reaction B is -75 kJ mol-1. Which reaction is faster? How do you know?
Nothing can be said about the rates of these two reactions. G is a state function and does not depend on the reaction rate.
A3.Why are condensed phases such as solids and pure liquids not included in an expression for an equilibrium constant? The concentrations of these species do not change as they are used up or created in a reaction – both the quantity and the volume change, but concentration is the quotient of the two, which does not change.
A4. In a reaction mechanism, what is the difference between a reaction intermediate and a catalyst? Define the term “rate limiting step.”
An intermediate is produced in an early step and consumed in a later step. A catalyst is consumed in an early step and reproduced in a later step. The rate limiting step is the slowest step in the mechanism of a multistep reaction.
A5.Which direction (left or right) will the equilibrium reaction CO(g) + ½ O2(g) ↔ CO2(g) shift if we do the following: Change| Direction that the equilibrium will shift
| Left| Right|
Lower the temperature| | |
Add some O2(g)| | |
Remove some CO2(g)| | |
Increase the pressure| | |
Dilute the system by adding some He(g)| | |
Increase the volume of the container| | |
Note the word “Dilute” in the last question. This means that the partial pressures of all gases are decreased. Thus Q>Keq, since the denominator in Q decreases faster than the numerator. Thus the left shift.
A6.Predict the signs of ΔHo and ΔSo, and under what conditions the reaction is spontaneous: Process| ΔHo (Check One!)| ΔSo (Check One!)| Is the reaction spontaneous below some T, above some T, at all T, or never?|
| Positive| Negative| Positive| Negative| |
H2O(g) → H2O(l)| | | | | Below|
NaCl(s) → Na+(g) + Cl-(g)| | | | | Above|
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)| | | | | Below| CaBr2(s) Ca+2(aq) + 2 Br-(aq)
(The solution is observed to get hot)| | | | | At all T| 16 CO2(g) + 18 H2O(g) → 2 C8H18(g) + 25 O2(g)| | | | | Never|
Part B: Long Answer Section: (10 marks each)
B1.Octane (C8H18) has a vapour pressure of 0.462 atm at 100 °C, and an enthalpy of vapourization of 50.1 kJ/mol. Calculate the normal boiling point of octane (oC).
Here, T1 = 100oC = 373 K
p1 = 0.462 atm
T2 = ?
p2 = 1.00 atm (since T2 is the normal boiling point,
ΔHvap = 50100 J mol–1
(b)Humans perspire as a way of keeping their bodies from overheating during strenuous exercise. The evaporation of perspiration transfers heat from the body to the surrounding atmosphere (ie skin). Calculate the total ∆S for evaporation of 1.0 g of water if the skin is at 37.5 °C and air temperature is 23.5 °C. (heat of vaporization of water is 40.79 kJ/mol)
The actual process involved can...