The concentration of NO 2 after 2 .5×10 2 sec and half-life period has to be calculated. Concept introduction: Integrated rate law for second order reactions: Taking in the example of following reaction, aA → products And the reaction follows second order rate law, Then the relationship between the concentration of A and time can be mathematically expressed as, 1 [ A ] t = kt+ 1 [ A ] 0 The above expression is called as integrated rate for second order reactions. Half life for second order reactions: In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A). The half-life of second order reaction can be calculated using the equation, t 1/2 = 1 (k [ A ] 0 ) Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.
The concentration of NO 2 after 2 .5×10 2 sec and half-life period has to be calculated. Concept introduction: Integrated rate law for second order reactions: Taking in the example of following reaction, aA → products And the reaction follows second order rate law, Then the relationship between the concentration of A and time can be mathematically expressed as, 1 [ A ] t = kt+ 1 [ A ] 0 The above expression is called as integrated rate for second order reactions. Half life for second order reactions: In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A). The half-life of second order reaction can be calculated using the equation, t 1/2 = 1 (k [ A ] 0 ) Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.
The concentration of NO2 after 2.5×102sec = 4.7×10-3M.
To calculate the half life of the reaction
The half-life of second order reaction can be calculated using the equation,
t1/2=1(k[A]0)
Given,
Concentration of NO2(A)=0.050M
Rate constant = 0.775L/(mol.s)
Then, the half life period is calculated as,
t1/2=1(0.755L(mol.s))(0.050mol/L)t1/2=25.80=26s
The half-life period of the reaction = 26s.
Conclusion
The concentration of NO2 after 2.5×102sec and half-life period was calculated using the integrated law and half-life period for second order reactions and were found to be 4.7×10-3M and 26s.
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b) Certain cyclic compounds are known to be conformationally similar to carbohydrates, although they are not
themselves carbohydrates. One example is Compound C shown below, which could be imagined as adopting
four possible conformations. In reality, however, only one of these is particularly stable. Circle the conformation
you expect to be the most stable, and provide an explanation to justify your choice. For your explanation to be
both convincing and correct, it must contain not only words, but also "cartoon" orbital drawings contrasting the
four structures.
Compound C
Possible conformations (circle one):
Дет
Lab Data
The distance entered is out of the expected range.
Check your calculations and conversion factors.
Verify your distance. Will the gas cloud be closer to the cotton ball with HCI or NH3?
Did you report your data to the correct number of significant figures?
- X
Experimental Set-up
HCI-NH3
NH3-HCI
Longer Tube
Time elapsed (min)
5 (exact)
5 (exact)
Distance between cotton balls (cm)
24.30
24.40
Distance to cloud (cm)
9.70
14.16
Distance traveled by HCI (cm)
9.70
9.80
Distance traveled by NH3 (cm)
14.60
14.50
Diffusion rate of HCI (cm/hr)
116
118
Diffusion rate of NH3 (cm/hr)
175.2
175.2
How to measure distance and calculate rate
For the titration of a divalent metal ion (M2+) with EDTA, the stoichiometry of the reaction is typically:
1:1 (one mole of EDTA per mole of metal ion)
2:1 (two moles of EDTA per mole of metal ion)
1:2 (one mole of EDTA per two moles of metal ion)
None of the above
Chapter 13 Solutions
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