Extra Credit 29
- Page ID
- 49188
Q 2.55
To what temperature must Li atoms be cooled so that they have the same Vrms as Cl2 at 25C?
Q 2.101
Calculate the mean kinetic energy (Etrans) in joules of the following molecules at 375 K-
H2
-
O2
-
NH3
Explain your results
Q 9.23
The rate law for the decomposition of hydrogen peroxide:2H2O2 (aq) 2H2O (l) + O2(g) with the rate = k[H2O2][I-]
The mechanism proposed for this process is
H2O2 + I- H2O + OI-
step 2 H2O2 + OI- H2O + O2 + I-
derive the rate law from these elementary steps. Clearly state the assumptions you use in the derivation.
Explain why the rate increases with increasing I- concentration.
Q 10.15
Calculate the concentration of a non competitive inhibitor (KI=3.85x10^2 M) needed to yield 75% inhibition of an enzyme catalyzed reaction.
Q 11.13
The Balmer series a series of spectral emission lines of the hydrogen atom that result from electron transitions from higher levels down to the energy level. The lyman series is a series of transitions as an electron goes from a high energy level to the lowest energy level of the electron. Calculate the shortest wavelength associated with the lymans series and the longest wavelength associated with the balmer series (in nm).
Q12.4
Ozone (O3) has a small dipole moment of (m=0.53D). Given that all the atoms in ozone explain how it has the ability to have a small dipole in terms of resonance.
Q13.8
According to Coulomb’s Law the electrical force (F) is inversely proportional to the square of the distance separating the two atoms involved. On the other hand van der waals forces are dependent on the short range repulsion.
-
Plot a general plot of the coulombic law’s effect of two atoms.
-
Based on the graph, provide a reason to why the distance of the atoms has a large dip and then seems to level out but not as largely as before.
Q2.68
Calculate the value of the most probable speed (Cmp) for CH4 at 20C. What is the ratio of the number of molecules with a speed of 876 m s^-1 to the number of molecules with this value of Cmp?Q14.11
The molar absorptivity of a potassium chloride solution is 4.2x10^2 L mol^-1 cm^-1 at 725 nm. The cell path is 2 cm and the concentration of KCl is 0.0059M. Calculate the absorbance of the solution at 725 nm.
Q9.7
The first order rate constant for the decomposition of a solution is 5.7x10^-4 s^-1 at 25C. The reaction is occurring in costant volume flask. Initially the pressure is 0.428 atm. What is the pressure if the system after 9.0 min? Assume ideal gas behavior.
Solutions:
Q.2.55
<u>rms= square root 3RT/M
Need to equate <u>rms for Li and Cl2
<u>rms(of Li) = <u>rms(of Cl2)
Square root 3RT(Li)/M(Li) = square root 3RT(of Cl2)/M(of Cl2)
T(of Li)=T(of Cl2)(M(of Li)/M(of Cl2))
T(of Li)=(298K)(6.94 g mol ^-1/70.9 g mol^-1)=29.2 K
Q.2.101
Etrans=3/2kbT=3/2(1.381x10^-23 J K^-1)(375K) = 7.768x10^-21
Therefore all a,b,c should equal to 7.768x10^-21 because the kinetic theory is statistical. This equation then states the when two ideal gases are at the same temperature they must have the same average kinetic energy. <Etrans> is independent of M of gas or any other properties except T.
Q.9.23
We can assume that k2>>k1, therefore the rate of the slowest elementary step should become the rate of the reaction. Which would be rate = k[H2O2][I-]
The rate should increase with increasing I- concentration because I- in this mechanism works as a catalysts, which is used to speed up the reaction.
Q.10.15
V0/(v0)inhibition = 1 + [I]/KI
0.25/0.75= 1 + [I]/ (3.85x10^2M)
(0.333)(3.85x10^2 M) – 1 = [I]
[I]= 1.26x10^2M
Q.11.13
The shortest wavelength in lyman series is n= (\infty)
Since the common spectral series of hydrogen in the lyman series n’=1 must substitute 2 with 1
1/ (\lambda) = R(1/(1^2) - 1/ (n^2)) = (1.097x10^7 m^-1)(1/ 1^2 - 1/ (\infty)^2)
(\lambda) = 91.16x10^-9 m = 91.16 nm
The longest wavelength in balmer series is n=3.
1/ (\lambda) = R(1/(2^2) - 1/ (n^2)) = (1.097x10^7 m^-1)(1/ 2^2 - 1/ 3^2)
(\lambda) = 653.3x10^-9 m = 653.3 nm
Q.12.4
Solution: Since the molecule has three oxygen atom in the lewis structure the bonds are different. Due to the resonance of the structure the bonds are able to change equaling to a slight dipole among the atoms. The v-shape of the molecules helps strengthen the dipole.
Q.13.8
Solution:
b. The atoms are attracted to each other up to a certain point which is the dip in the graph. Then the atoms are either really repelled by each other with the distance of them being to close. The graph levels out to the left because that is the point of which the atoms are attracted to each other.
Q.14.11
A= (epsilon)bc
A= (absorptivity)(path length)(concentration)
A = (4.2x10^2 L mol^-1 cm^-1) (2 cm) (0.0059M) = 4.956R= 8.314 J K^-1 mol^-1 T=293K
Q.2.68
M= 0.01604 kg/mol
Cmp = square root of 2RT/M
= square root of 2x8.314 J K^-1 mol^-1 x 293K/0.01604 kg/mol = 303740.90 J kg^-1
= square root of 3.04 x 10^5 m^2 s^-2
=551 m s^-1
Ratio of 800:551
Q.9.7
ln[A] = -kt+ln[A]0 PV=nRT
[A]= n/V=P/RT
[A]/[A]0=e^-kt
P/RT/(0.428/RT)=e^-(5.7x10^-4 s^-1)(540s)
P/0.428=1.360
P=0.582 atm