Extra Credit 34

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Q. 2.63

If the speeds of 10 particles (in cm s^-1) are found to be 0.2, 1.0, 1.4, 1.4, 2.0, 2.5, 2.7, 3.0, 3.4, and 4.2, how would you calculate:

(a) the average speed

(b) the root-mean-square speed

(c) the most probable speed of these particles

Explain.

Q. 2.72

Calculate the mean free path and the binary number of collisions (L^-1s^-1) between HI molecules at room temperature and 2.5 atm. The collision diameter of HI may be assumed to be 5.10 Å. Assume ideal-gas behavior.

Q. 9.11

What is the half life of a compound if 75% of a given sample of the compound decomposes in 60 min? Assume first-order kinetics. How would the half life differ if we assumed zero-order kinetics? Second-order?

Q. 9.4

The following reaction is found to be first order in A:

A --> B + C

If half of the starting quantity of A has been depleted after 48 seconds, what is the fraction that will be depleted after 6.0 min?

Q. 9.28

Over a range of about \pm3 ºC from what is considered to be normal body temperature, the metabolic rate, M_T, is given by M_T= M₃₇(1.1)^ΔT, where M₃₇ is the normal rate and ΔT is the change in T. How would you describe this equation in terms of a possible molecular interpretation?

Q. 10.3

The enzyme acetylcholinesterase catalyzes the hydrolysis of acetylcholine. If it has a turnover rate of 25,000 s^-1, then how long does it take for the enzyme to cleave one acetylcholine molecule?

Q. 11.18

Ozone (O₃) in the stratosphere absorbs harmful radiation from the sun by undergoing decomposition:

(a) Referring to Appendix 2, calculate the Δ_rHº value for this process

(b) What is the maximum wavelength of photons (in nm) that possess this energy in order for the decomposition of ozone to occur photochemically?

Q. 12.12

How would you use the molecular orbital theory to describe the bonding scheme in the following species: H₂+, H₂, He₂+, and He₂? Arrange the species in order of increasing stability.

Q. 13.13

Explain the difference in the boiling points of diethyl ether (C₂H₅OC₂H₅) and 1-butanol (C₄H₉OH). The former has a boiling point of 34.5ºC whereas the latter boils at 117ºC. If the two compounds have the same type and number of atoms then why the huge discrepancy?

Q.14.16

The equilibrium bond length in nitric oxide (¹⁴N¹⁶O) is 1.15 Å (1 Å = 1x10¹⁰m). Calculate:

(a) the moment of inertia of NO

(b) the energy for the J = 0 --> 1 transition.

(c) the number of times the molecule rotates per second at the J = 1 level

S. 2.63

(a) The average speed:

v_avg= \(\dfrac{0.2+1.0+1.4+1.4+2.0+2.5+2.7+3.0+3.4+4.2}{10}\) = 2.18cm s^-1

(b) The root-mean-square speed:

v_rms = \(\sqrt{\dfrac{0.2^2+1.0^2+1.4^2+1.4^2+2.0^2+2.5^2+2.7^2+3.0^2+3.4^2+4.2^2}{10}}\) = 2.46cm s^-1

(c) The most probable speed:

v_mp = 1.4cm s^-1 because that's the mode

S. 2.72

Useful information:

mean free path: \(\lambda\) = \(\dfrac{1}{\sqrt{2}\pi(d^2)(\dfrac{N}{V})}\)

ideal gas law: PV = nRT or V = \(\dfrac{nRT}{P}\)

binary number of collisions: Z_ii = \(\dfrac{\sqrt{2}}{2}\pi(d^2)\)<c>\((\dfrac{N}{V})^2\), where <c> = \((\dfrac{8k_BT}{\pi m})\)^1/2

T = 298K
P = 2.5atm
d = 5.1x10^-10m
N = 6.022x10²³, assuming 1 mole present

To find the mean free path:

\(\lambda\) = \(\dfrac{1}{\sqrt{2}\pi(d^2)(\dfrac{N}{V})}\)

V = \(\dfrac{nRT}{P}\) = 9.78L

\(\lambda\) = \(\dfrac{1}{\sqrt{2}\pi(d^2)(\dfrac{N}{V})}\) = 1.406x10^-5m

To find the binary number of collisions:

Z_ii = \(\dfrac{\sqrt{2}}{2}\pi(d^2)\)<c>\((\dfrac{N}{V})^2\) = 7.42x10^-50L^-1s^-1

S. 9.11

The half life for 1st order kinetics is constant. If we set up a table with known information, we can solve for the half life.

	[C] (%)	t (min)
1	-	-
2	-	-
3	25	60

Since the half life for 1st order kinetics is constant:

	[C] (%)	t (min)
1	100	0
2	50	30
3	25	60

The half life is 30min.

1st order kinetics do not depend on the original concentration, but 0th order and 2nd order kinetics do depend on the original concentration. Therefore, we can solve 1st order kinetics without a given original concentration, but we cannot do the same for 0th order and 2nd order kinetics.

S. 9.4

The half life for 1st order kinetics is constant. If we set up a table with known information and fill in the rest, we can solve for the fraction that will be depleted after 6.0 min.

	[C] (%)	t (sec)
1	100	0
2	50	48
3	25	96
4	12.5	192
5	6.25	384

And since 6min = 360sec, we can look at the table and determine the closest value at that time. The closest thing to 360sec is 384sec, so the fraction that will be depleted after 6min is around \(\dfrac{1}{5}\)

S. 9.28

Molecularly, this metabolic rate is affected by temperature. This equation is solved by using the normal metabolic rate (M₃₇) and multiplying it by 1.1^ΔT, where ΔT is the change in temperature.

S. 10.3

Each acetylcholinesterase molecule can produce up to 25,000 molecules of acetylcholine per second.

If we set up a proportion:

\(\dfrac{1 acetylcholine}{x}\) = \(\dfrac{25,000 acetylcholine}{1sec}\)

x = 4x10^-5sec

S. 11.18

(a) To do this, we need the formula:

ΔHº_reaction = ∑ΔHº_f(products) - ∑ΔHº_f(reactants)

and we need the reaction:

2O₃(g) → 3O₂(g)

From Appendix 2, we know:

Δ_rHº of O₃ = 142.7kJ/mol
Δ_rHº of O2 = 0kJ/mol

Now, we can solve for everything.

ΔHº_reaction = ∑ΔHº_f(products) - ∑ΔHº_f(reactants) = 3(0kJ/mol) - 2(142.7kJ/mol) = -284.4kJ/mol

(b) Here, we use the equation:

E = \(\dfrac{hc}{\lambda}\)

and if we rearrange it to solve for \(\lambda\):

\(\lambda\) = \(\dfrac{hc}{E}\) = \(\dfrac{(6.626x10^{-31}Js)(3x10^8m/s)}{-284400J}\) = 6.989x10^-31m

S. 12.12

Bond order equation:

BO = (1/2)(#bonding orbitals - #antibonding orbitals)

BO of H₂⁺: (1/2)(1-0) = 1/2

BO of H₂: (1/2)(2-0) = 1

BO of He₂⁺: (1/2)(2-1) = 1/2

BO of He₂: (1/2)(2-2) = 0

In order of increasing stability: He₂ < He₂⁺ < H₂+ < H₂

S. 13.13

1-butanol has the -OH group at one end, so electronegativity is stronger on that side of the molecule, thus increasing its dipole moment. Diethyl ether has its electronegative Oxygen in the middle of the molecule. The placement of an electronegative atom within a molecule really affects its dipole moment and its boiling point, along with other bond-related properties.

S.14.16

(a) The equation for moment of inertia is:

I = m₁r₁² + m₂r₂²

where:

m_{1 =}¹⁴N = 14.01 g/mol
m_{2 =}¹⁶O = 16 g/mol
r₁ and r₂ = \(\dfrac{1.15x10^{-10}m}{2}\) = 5.75x10^-11m

I = m₁r₁² + m₂r₂² = (14.01 g/mol)(5.75x10^-11m)² + (16 g/mol)(5.75x10^-11m)² = 9.922x10^-20g*m²*mol^-1

(b) To find this, we use:

ΔE = E₁ - E₀ = 2Bh, where B = \(\dfrac{h}{8\pi^2 I}\)

so:

ΔE = E₁ - E₀ = \(\dfrac{2h^2}{8\pi^2 I}\) = \(\dfrac{2(6.626x10^{-34}Js)^2}{8\pi^2 (9.922x10^{-20}gm^2mol^{-1})}\) = 1.12x10^-49J

(c) Use the equation:

E_rot = \(\dfrac{J(J+1)h^2}{8\pi^2I}\)

E_rot = \(\dfrac{1(1+1)(6.626x10^{-34}Js)^2}{8\pi^2 (9.922x10^{-20}gm^2mol^{-1})}\) = 0 J