Short Event Description: Teams will demonstrate chemistry laboratory skills related to kinetics and gases.
The rate law for the following reaction is: .
Rank the reaction mixtures below from fastest to slowest at a given temperature.
Re: Chemistry Lab C
Posted: August 31st, 2015, 5:42 am
by John Richardsim
bernard wrote:Short Event Description: Teams will demonstrate chemistry laboratory skills related to kinetics, chemical reactions and stoichiometry.
(Psssst, the short event descriptions are not updated (new ones have been added and old ones removed, but for events with changing topics, the description was not changed for the new topic (e.g., cardiovascular and immune are listed as the A&P body systems and the protein modeling description still says "2015" in it). Stoichiometry has been replaced with gas laws.)
Re: Chemistry Lab C
Posted: August 31st, 2015, 9:00 am
by bernard
Fixed it with the new description.
Re: Chemistry Lab C
Posted: September 1st, 2015, 12:23 pm
by samlan16
B, D, A, C
Re: Chemistry Lab C
Posted: September 1st, 2015, 12:45 pm
by bernard
Correct! From fastest to slowest the reaction mixtures are: B > D > A > C. From the rate law, we know that the rate is proportional to . Since the reaction mixtures are all the same volume, the concentrations of each reactant are proportional to the number of reactant molecules. Therefore, the rate of B ∝ (4)²(2) = 32, D ∝ 18, A ∝ 16 and mixture C will have rate of zero since it doesn’t have any chlorine.
Re: Chemistry Lab C
Posted: September 25th, 2015, 10:20 am
by finagle29
Starting this up again as it's been more than three weeks.
Question: For the following types of processes, name the property of a gas undergoing such a process that would be constant: isochoric, adiabatic, isobaric.
Re: Chemistry Lab C
Posted: December 21st, 2015, 3:13 pm
by arvind_r
It's been a while since there was activity on this thread.
Yes
I'll give it to you. I meant to say reversible adiabatic and was looking for entropy, but without reversible, entropy is not always constant, so my "more right" answer isn't actually more right.
Yes
EDIT: oops I forgot to say it outright, but it's your turn now, arvind_r.
Re: Chemistry Lab C
Posted: January 1st, 2016, 6:17 pm
by arvind_r
A question from the National Chemistry Olympiad with some modifications:
Consider the gas-phase reaction between nitric oxide and oxygen showing the initial concentrations of the reactants at a
temperature, 300 K:
Experiment [NO], [O2], Initial Rate of NO2 formation,
1 0.020 0.020 0.057
2 0.040 0.040 0.455
3 0.040 0.020 0.228
a. Determine the order with respect to .
b. Determine the order with respect to .
c. Calculate the rate constant and give its units at this temperature.
d. If this reaction follows a two-step mechanism with the first step being
i. Write an equation for the second step of the mechanism.
ii. Identify the rate determining step of this mechanism and outline your reasoning.
e.This experiment is repeated at a temperature of 400K. The rate constant is found to be double the rate constant observed at 300K. Find for the reaction.
The decomposition of ozone in the reaction 2 O3 (g) → 3 O2 (g) has the following proposed
mechanism:
Step 1: O3 ⇄ O2 + O
Step 2: O + O3 ⇄ O2 + O2
Determine the rate law for the decomposition of O3, given that the slow step is the
second step and the reverse reaction of step 2 is so slow that it can be ignored
Re: Chemistry Lab C
Posted: January 5th, 2016, 7:25 pm
by finagle29
a. The reaction is 2nd order w.r.t. [math]NO[/math]
b. The reaction is 1st order w.r.t. [math]O_2[/math]
c. The rate constant [math]k=7110M^{-2}s^{-1}[/math]
d. i. [math]O_2 + N_2O_2 \longrightarrow 2 NO_2[/math]
ii. The rate determining step is the second one. It suggests a rate law Rate=[math]k[NO]^2[O_2][/math], which is consistent with the data.
e. [math]E_a=6920 Jmol^{-1}[/math]
Without looking at the answer key, the answer should be similar to the one used to determine which step is the slow step in part d of your question.
Step 1 goes to equilibrium quickly so that Rate_forwards = Rate_backwards = k_1 [O3] = k_-1 [O2][O]
Step 2 has rate law Rate = k_2 [O][O3]
solving for [O] in the first equation and substituting into the second yields
Rate = k_1 k_2 / k_-1 [O3]^2 / [O2] or just Rate = k [O3]^2 / [O2]
Does this help, or should I explain further?