Question

Some measurements of the initial rate of a certain reaction are given in the table below.

N2 |
H2 |
initial rate of reaction |
---|---|---|

1.08 |
1.38 |
0.720/ |

1.08 |
1.93 |
1.41/ |

3.76 |
1.38 |
8.73/ |

Use this information to write a rate law for this reaction, and calculate the value of the rate constant

*k*.Be sure your value for the rate
constant has the correct number of significant digits. Also be sure
your answer has the correct unit symbol.

Answer #1

At a certain temperature the rate of this reaction is first
order in HI with a rate constant of :0.0632s
2HIg=H2g+I2g Suppose a vessel contains HI at a concentration of
1.28M . Calculate how long it takes for the concentration of HI to
decrease to 17.0% of its initial value. You may assume no other
reaction is important. Round your answer to 2 significant
digits.

Constants | Periodic Table
The following reaction was monitored as a function of
time:
A→B+C
A plot of ln[A] versus time yields a straight
line with slope −5.0×10−3 /s .
Part A
Part complete
What is the value of the rate constant (k) for this
reaction at this temperature?
Express your answer using two significant figures.
k =
5.0×10−3
s−1
Previous Answers
Correct
Part B
Part complete
Write the rate law for the reaction.
Rate=k
Rate=k[A]
Rate=k[A]2
Rate=k[A]3
Previous Answers...

Please study the table of data below, collected for the gas
phase combination reaction of N2O with O3 at 298 K and constant
volume (1.00 L).
Experiment
[N2O]0
[O3]0
Initial rate of reaction of O3
(M/s)
1
.010
.010
6.01 x 10^-4
2
.010
.020
1.20 x 10^-3
3
.020
.020
1.19 x 10^-3
A) Please determine the order of each reactant and write the rate
law for the overall reaction.
B) The overall reaction is: N2O (g) + O3...

The rate constant for a certain reaction is k = 2.00×10−3 s−1 .
If the initial reactant concentration was 0.400 M, what will the
concentration be after 20.0 minutes? Express your answer with the
appropriate units.
Part B A zero-order reaction has
a constant rate of 4.70×10−4 M/s. If after 70.0 seconds the
concentration has dropped to 2.00×10−2 M, what was the initial
concentration?

The rate constant for a certain reaction is k = 7.20×10−3 s−1 .
If the initial reactant concentration was 0.850 M, what will the
concentration be after 10.0 minutes?
A zero-order reaction has a constant rate of
3.00×10−4M/s. If after 60.0 seconds the
concentration has dropped to 3.50×10−2M, what
was the initial concentration?

3. A certain reaction has the following general form: 2A > B
Concentration vs time data were collected for this reaction, at
50°C and an initial concentration of 0.0200 M. It is determined
that a plot of ln[A] vs. time resulted in a straight line with a
slope value of - 2.97 X 10-2 min-1.
A. Write the rate law.
B. Write the integrated rate law
C. What is k for this reaction (or what is the rate constant for...

The following initial rate data are for the reaction of ammonium
ion with nitrite ion in aqueous solution: NH4+ + NO2- N2 + 2 H2O
Experiment [NH4+]o, M [NO2-]o, M Initial Rate, Ms-1
1. 0.296 0.180 1.85×10-5
2. 0.296 0.361 3.72×10-5
3. 0.592 0.180 3.71×10-5
4. 0.592 0.361 7.44×10-5
Complete the rate law for this reaction in the box below. Use
the form k[A]m[B]n , where '1' is understood for m or n and
concentrations taken to the zero power...

The rate constant for a certain reaction is k = 4.40×10−3 s−1 .
If the initial reactant concentration was 0.450 M, what will the
concentration be after 3.00 minutes? Express your answer with the
appropriate units

The integrated rate law allows chemists to predict the reactant
concentration after a certain amount of time, or the time it would
take for a certain concentration to be reached. The integrated rate
law for a first-order reaction is: [A]=[A]0e−kt Now say we are
particularly interested in the time it would take for the
concentration to become one-half of its initial value. Then we
could substitute [A]02 for [A] and rearrange the equation to:
t1/2=0.693k This equation calculates the time...

The integrated rate law allows chemists to predict the reactant
concentration after a certain amount of time, or the time it would
take for a certain concentration to be reached. The integrated rate
law for a first-order reaction is: [A]=[A]0e−kt Now say we are
particularly interested in the time it would take for the
concentration to become one-half of its inital value. Then we could
substitute [A]02 for [A] and rearrange the equation to: t1/2=0.693k
This equation caculates the time...

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