Modern family: Addition

State level Olympiad (8)

24. Which substance has a non-zero standard free energy of formation?

(A) Pb(s) (B) Hg(l) (C) Cl2(g) (D) O3(g)

Sol. The most stable elemental form of each element in their standard states have a value of 0 for standard free
energy of formation, so we need to determine the answer choice that is not in standard state or not the most stable
elemental form. We see that O3 is not the most stable elemental form of oxygen, as O2 is the standard state of
oxygen. Thus, the answer is D .

25. Factors that can affect the rate of a chemical reaction
between a solid and a solution include all of the
following EXCEPT the

(A) concentration of the reactants in solution.

(B) volume of the container.

(C) size of the solid particles.

(D) temperature.

Sol. Both temperature and concentration are factors that can increase the rate of the reaction. In addition, in-
creasing the size of the solid particles will decrease the rate, as the surface area is smaller and less particles are exposed. Thus, the only non contributing factor is volume of the container , or B . This is because the reaction
involves a solution reacting with a solid, so change in volume of the container doesn’t affect the surface area of the
solid nor the concentration of the solution.

26. The commercial production of ammonia is represented by
the equation N2(g) + 3H2(g) → 2NH3(g). If the rate of disappearance of H2(g) is 1.2 × 10-3 mol/min, what is the rate of appearance of NH3(g)?

(A) 2.4 × 10-3 mol/min
(B) 1.8 × 10-3 mol/min
(C) 1.2 × 10-3 mol/min
(D) 8.0 × 10-4 mol/min

Sol. We can do dimensional analysis to determine the rate of NH3.
(1.2 × 10−3 mol H2 /min) ×
(2 mol NH3 / 3 mol H2)
= 8.0 × 10−4 mol NH3 /min
Thus, the correct answer is D .

27. A reaction is endothermic with ∆H = 100. kJ/mol.
If the activation enthalpy of the forward reaction is 140.
kJ/mol, what is the activation enthalpy of the reverse
reaction?

(A) 40. kJ/mol
(B) 100. kJ/mol
(C) 140. kJ/mol
(D) 240. kJ/mol

Sol. With an energy profile diagram (energy versus reaction progress) sketched, we can easily find that the relationship between the enthalpy change and the activation energy of forward and reverse reaction.
∆H = E_a (f) − E_a (r)
where E_a (f) = 140 kJ is the forward rate activation energy, and E_a (r) is the reverse rate activation energy. We are
given that ∆H = 100 kJ, so we can just substitute to solve for
E_a (r).
140 kJ = E_a (r) + 100 kJ
E_a (r) = 40 kJ
Thus, the correct answer is A .

28. The first-order disappearance of a substance has a half-
life of 34.0 s. How long does it take for the concentration
of that substance to fall to 12.5% of its initial value?

(A) 11 s
(B) 68 s
(C) 102 s
(D) 272 s

Sol. First we can use the first-order half-life equation and solve for k.
t_(1/2) = ln(2)/k

34.0 s = ln(2)/k

k = (ln 2/34.0) s^−1

Now, to solve for the time when [A] = (1/8) [A_0], we can substitute our value of k into the first-order integrated rate law.
ln[A] = −kt + ln[A_0]
kt = ln[A_0] − ln[A] =
ln{[A_0]/[A]} = ln 8

t = ln(8) / k
= 34.0 ln(8) / ln(2) = 34.0 × (log8 base 2)= 34.0 × 3 = 102 s
Thus, the answer is C . Alternatively, we can simply recognize that a “one-eighth life” is three half-lives because
(1/2)^3 = 1/8.

29. The reaction C3H7I + Cl-
→ C3H7Cl + I-
is thought to occur in
the polar solvent CH3OH by the mechanism:
Step 1: C3H7I → C3H7+
+ I-
(slow)
Step 2: C3H7+
+ Cl-
→ C3H7 Cl (fast)

Which species is an intermediate in this reaction?

(A) CH3OH
(B) C3H7+
(C) I-
(D) Cl-

Sol. The definition of an intermediate is a substance that is formed in one step, and consumed in a later step.
The substance that fits this definition is C3H7+ , or B , which is produced in Step 1 and consumed in Step 2.

30. The times listed are those recorded at 25˚C for the reaction
below to produce a measurable amount of I2(aq):
(S2O8)2-(aq) + 2I-
(aq) → I2(aq) + 2(SO4)2-(aq)
Experiment Initial
[(S2O8)2-], M
Initial
[I-], M
Time
(sec)

1 0.0400 0.0800 39
2 0.0400 0.0400 78
3 0.0100 0.0800 156
4 0.0200 0.0200 ?

What is the expected time for experiment 4?

(A) 156 sec
(B) 234 sec
(C) 312 sec
(D) 624 sec

Sol. First we must determine the rate law. From experiment 1 to experiment 2, [I−] was divided by 2 and the rate was divided by 2 (thus the amount of time was multiplied by 2). Therefore, the reaction is first order in I−. From
experiment 1 to experiment 3, [S2O8^2−] was divided by 4 and the rate was divided by 4. Therefore, the reaction is
first order in S2O8^2−.
Finally, from reaction 3 to reaction 4, we first multiply the rate by 2 because [I−] was multiplied by 2. Then we
divide the rate by 4 because [S2O8^2−] was divided by 4. So far we have divided the rate by 2. Since the time is
inversely proportional to rate, we multiply the time by 2. Therefore, the correct answer is C .
T4 = T3 · 2 = 156 × 2 = 312 sec

31. What is the equilibrium expression, Kc, for the reaction:
2S(s) + 3O2(g) 2SO3(g)?

(A) Kc = 2[SO3]/(2[S]+3[O2])
(B) Kc = 2[SO3]/3[O2]
(C) Kc = [SO3]^2/[S]^2[O2]^3
(D) Kc = [SO3]^2/[O2]^3

Sol. Note that S(s) will not count in calculating the equilibrium expression, as it is not aqueous or gaseous. (Its concentration is constant.) Therefore, the correct answer is D .
Kc = [SO3]^2 / [O2]^3

32. In which reaction at equilibrium will the amount of
reactants present increase with an increase in the
container volume?

(A) C(s) + CO2(g) 2CO(g)
(B) H2(g) + F2(g) 2 HF(g)
(C) CO(g) + NO2(g) CO2(g) + NO(g)
(D) N2(g) + 3 H2(g) 2NH3(g)

Sol. Let us account for both an increase and decrease in volume. When the volume is increased, pressure decreases.
The equilibrium state will shift to favor the side of the reaction with more moles of gases (thus increasing pressure).
Similarly, a decrease in volume will shift the equilibrium state to favor the reaction with fewer moles. Looking at all
of the answer choices, only the reaction below has more moles on the reactants than the products, and will therefore
increase in reactant moles with a volume increase.
N2(g) + 3H2(g) -b-o-t-h- -w-a-y-s- 2NH3(g)
Thus, the correct answer is D .

Similarly remaining questions were from organic chemistry reactions and isomers which are not very easy to present as matter of drawing.