How do I solve an AP chem free response on ammonia and light interactions?

In summary, the conversation is about solving a chemistry free response problem on the equilibrium of ammonia in aqueous solution and calculating various values such as equilibrium constant, pH, base ionization constant, and percent ionization. The second problem involves a rate study of a reaction and determining the order of the reaction, rate law, and specific rate constant. The conversation also includes a discussion on the interactions of light with molecules, atoms, and ions and calculations involving frequency, energy, and bond energy.
  • #1
doremi
1
0
[SOLVED] an AP chem free response (please help!)

i have a final tomorrow (yea i was procrasinating) and i have to figure out how to do these practice problems! please help you can! i need to know how to do it ASAP!

NH3(aq) + H20(l) <---> NH4+(aq) + OH-(aq)

In aqueous solution, ammonia reacts as represented above. In 0.0180M NH3(aq) at 25C, the hydroxide ion concentration, [OH-], is 5.60x10^-4M. In answering the following, assume that temperature is constant at 25C and that volumes are additive.

a) Write teh equilibrium-constant expression for the reaction represented above.
b) Determine the pH of 0.0180M NH3(aq).
c) Determine the value of the base ionization constant, Kb, for NH3(aq).
d) Determine the percent ionization of NH3 in 0.0180M NH3 (aq).
e) In an experiment, a 20.0 mL sample of 0.0180M NH3 (aq) was placed in a flask and titrated to the equivalence point and beyond using 0.0120 M Hcl(aq).
i) Determine the volume of 0.0120 M HCl (aq) that was added to reach teh equivalence point.
ii) Determine the pH of the solution in teh flask after a total of 15.0mL of 0.0120 M Hcl(aq) was added.
iii) Determine the pH of the solutiobn in the flask after a total of 40.0mL of 0.0120M Hcl(aq) was added.

Answer the following questions regarding light and its interactions with molecules, atoms, and ions.
a) The longest wavelength of light with enough energy to brea the cl-cl bond in cl2(g) is 495nm.
i)Calculate the frequency, in s^-1, of the light.
ii) Calculate the energy, in J, of a photon of the light.
iii) Calculate the minimum energy, in kJ mol^-1, of the Cl-Cl bond.
b) A certain line in teh spectrum of atomic hydrogen is associated with electronic transition in the H atom from the sixth energy level (n=6) to the second energy level (n=2).
i) Indicate whether the H atom emits energy or whether it absorbs energy during the trasition. Justify your answer.
ii) Calculate the wavelength, in nm, of the radiation associated with teh spectral line.
iii) Account for the observation that the amount of energy associated with the same elctronic transition (n=6 to n=2) in the He+ ion is greater than that associated with the corresponding transition in the H atom.





and this is the second problems

2NO(g) + Br2(g) ---> 2NOBr(g)

A rate study of teh reaction represented was conducted at 25C. The data that were obtained are shown in the table below.



Experiment Initial [NO] Initial [Br2] Initial Rate of Appearance
(mol L^-1) (mol L^-1) of NOBr (mol L^-1s^-1)
1 0.0160 0.0120 3.24x10^-4
2 0.0160 0.0240 6.38x10^-4
3 0.0320 0.0060 6.42x10^-4







a) Calculate the initial rate of dissapearance of Br2(g) in experiment 1.
b) Determine the order of the reaction with respect to each reactant, Br2(g) and NO(g). In each case, explain your resoning.
c) For the reaction,
i) Write the rate law that is consitent with the data, and
ii) calculate the value of the specific rate constant k, and specify units.
d) the following mechanism was proposed for the reaction:

Br2(g)+NO(g) ---> NOBr2(g) slow
NOBr2(g) + NO(g) ---> 2NOBr(g) fast

In this mechanism consitent with the given experimental observations? Justify your answer.
 
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  • #2
For Question No 3
[tex] Rate=k[NO]^m[Br_2]^n[/tex]
where m and n are order of reaction w.r.t [NO]&[Br2]
Code:
Experiment   Initial[NO]    Initial[Br2]    Initial Rate of Appearance
              (mol L^-1)    (mol L^-1)      of NOBr (mol L^-1s^-1)
1            0.0160          0.0120            3.24x10^-4
2            0.0160          0.0240            6.38x10^-4
3            0.0320          0.0060            6.42x10^-4

so u see from the above equation u see if conc of NO remains constant on doubling the conc of bromine we see the rate doubles hence
[tex]2^n=2[/tex]

therefore n=1
Similarly for EXP 3 u have Rate=6.42*10^-4
and rate for EXP 1 =3.24*10^-4
divide Rate for Exp 3 by Exp 1 u get

[tex]2^m*2^{-1}=2[/tex]
so m=2
 
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  • #3
I forget to welcome u and i want to direct you to sticky https://www.physicsforums.com/showthread.php?s=&threadid=28

Though above post is a clear solution still i give u some hints
initial Rate of disappearance of Br2= initial rate of appearance of NOBr

[tex]-\frac{d[NO]}{2dt}=-\frac{d[Br_2]}{dt}=\frac{d[NOBr]}{dt}[/tex]
I hope u will caluate it for part c
 
  • #4


Originally posted by doremi

Answer the following questions regarding light and its interactions with molecules, atoms, and ions.
a) The longest wavelength of light with enough energy to brea the cl-cl bond in cl2(g) is 495nm.
i)Calculate the frequency, in s^-1, of the light.
ii) Calculate the energy, in J, of a photon of the light.
iii) Calculate the minimum energy, in kJ mol^-1, of the Cl-Cl bond.
b) A certain line in teh spectrum of atomic hydrogen is associated with electronic transition in the H atom from the sixth energy level (n=6) to the second energy level (n=2).
i) Indicate whether the H atom emits energy or whether it absorbs energy during the trasition. Justify your answer.
ii) Calculate the wavelength, in nm, of the radiation associated with teh spectral line.
iii) Account for the observation that the amount of energy associated with the same elctronic transition (n=6 to n=2) in the He+ ion is greater than that associated with the corresponding transition in the H atom.

i) [tex]f=\frac{c}{\lambda} \mbox{where c is speed of light }[/tex]
ii) E=hf where h planks constant
iii) [tex] E_{min}=\frac{hc}{\lambda}=hf[/tex]
B)
It will emit a photon or Energy
[tex]E_n=-\frac{-13.6eV}{n^2}[/tex] ... for Hydrogen atom
[tex]E_6-E_2=\frac{hc}{\lambda}[/tex]

Since energy is due to Electrostatics interaction And we observe that The nuclei of helium have different no. of Protons than that of hydrogen so the above energy equation has a factor of Z^2 in addition
 
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  • #5
njkvbj,m
 
  • #6
himanshu121 said:
B)
It will emit a photon or Energy
[tex]E_n=-\frac{-13.6eV}{n^2}[/tex] ... for Hydrogen atom

Is this from the equation:

[tex] E_n=-\frac{Cz^2}{n^2}[/tex]

I was taught this last year, the prof didnt tell us the value of C for hydrogenic atoms though :<
 

1. What is an AP chem free response?

An AP chem free response is a type of question that appears on the Advanced Placement Chemistry exam. It is a long-form question that requires students to apply their knowledge of chemistry concepts to solve a problem or answer a prompt. It is typically worth a significant portion of the overall exam score.

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The number of AP chem free response questions varies each year, but there are typically between 4-7 questions on the exam. The exact number will be specified on the exam instructions.

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Yes, scratch paper is allowed for the AP chem free response questions. However, all final answers must be written on the designated answer sheet provided by the exam proctor. Any work done on scratch paper will not be graded.

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