Extra Credit 55

2.58)

Which of the atoms will give the greatest at 30°C? a) He b) C c) O? Solve this problem by using the equation for but without calculation.

Solution:

c_rms = v_rms = \(\sqrt{\frac{3RT}{M}}\)

The smaller the molar mass, the greater the , so He has the greatest root-mean-square speed (C_rms).

9.2)

The rate law for the reaction:

is given by rate = k [NH₄⁺][NO₂^-]. At 25ºC, the rate constant is 3.0 x 10^-4 M^-1s^-1. Calculate the concentration of [NH₄⁺] if the rate is 7.23 x 10^-6 M s^-1 and [NO₂^-] = 0.03M.

Solution:

rate = k [NH₄⁺][NO₂^-]

7.23 x 10^-6 M s^-1 = (3.0 x 10^-4 M^-1s^-1)(0.03M)[NH₄⁺]

[NH₄⁺] = 0.8033M

9.26)

For reaction:

\[O_3 + O \rightarrow 2O_2\]

\[Rate= k{\frac{[O_3]^2}{[O_2]}}\]

Calculate the activation energy at 350K, 500K, and 1000K with the rate constant 3 x 10⁹S^-1. Assume A = 10¹⁰S^-1 in each case.

Solution:​

\[k=Ae^{\frac{-E_a}{(R)(T)}}\]

At 350K:

\[3·10^9=10^{10}e^{\frac{-E_a}{(8.314)(350K)}}\]

E_a = 3503 Jmol^-1

At 500K:

\[3·10^9=10^{10}e^{\frac{-E_a}{(8.314)(500K)}}\]

Ea = 5004 Jmol^-1

At 1000K:

\[3·10^9=10^{10}e^{\frac{-E_a}{(8.314)(1000K)}}\]

E_a = 10009 Jmol^-1

10.1)

What assumption is made differently in between the rapid equilibrium and the steady state kinetics?

Solution:

Rapid equilibrium assumes k_-1>>k₂, so the enzyme complex is dissociating faster than product. Steady-state kinetics assumes

k₂ >>k₁, so the enzyme complex is forming at the same rate as it is turning into product.

11.16)

The retina of a human eye can detect light when radiant energy incident on it is at least 4.0 x 10^-17J. For light of 320 nm, how many photons does this correspond to?

Solution:

\[E={\frac{hc}{\lambda}}\]

\[E={\frac{(6.626·10^{-34})(3·10^8)}{320·10^9}}\]

\[E=6.212·10^{-19}J\]

\[64 photons={\frac{4.0·10^{-17}}{6.21·10^{-19}}}\]

12.7)

Consider the following Lewis structure for . Does it satisfy the octet rule? If not, draw additional resonance structures that do satisfy the octet rule.

Solution:

It doesn't satisfy the octet rule because it is missing lone pairs of electrons on each oxyge. The following resonance structrues satisgy the octet rule:

13.11)

Calculate the bond enthalpy of KCl using the Born-Haber cycle. The bond length of KCl is 212pm. The first ionization energy is 418.KJ mol^-1 and the electron affinity for Cl is 349 KJ/mol. Use n = 10.

Solution:

\[V_o={\frac{q_+q_-}{4\pi\varepsilon_or_e}}(1-{\frac{1}{n}})\]

\[V_o={\frac{(1.602·10^{-19})^2(6.022·10^{23})}{4\pi(8.854·10^{-12})(212·10^{-12})}}(1-{\frac{1}{10}})\]

\[V_o=5.897·10^5N·m/mol\]

\[V_o=589.7KJ/mol\]

\[-D=I_1-E_A+V_o\]

\[-D=418.71-349+(-589.7)=-520KL/mol\]

14.4)

Convert the following percent transmittance to absorbency: a) 80% b) 35% c) 100%

Solution:

\[-log(T)=A\]

\[A)-log(0.80)=0.097\]

\[B)-log(0.35)=0.46\]

\[C)-log(1)=0\]

9.9)

The progress of a reaction in the aqueous phase was monitored by absorbance of a reactant at the following times:

time(s)	0	54	171	390	720	1010	1190
Absorbance	1.34	1.25	1.07	0.807	0.526	0.360	0.285

Assuming 1st order kinetics, find the rate constant.

Solution:

\[ln{\frac{[A]}{[A_o]}}=-kt\]

\[ln{\frac{[1.25]}{[1.34]}}=-k(54s)\]

\[k=0.0013s^{-1}\]

2.70)

How does the molecular collisions depend on a) the average speed b) the number of molecules c) the volume d) the separation between the 2 spheres?

Solution:

a) directly proportional

b) directly proportional

c)inversely proportional

d)directly proportional