Answer:

Explanation:
Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.
Hence, the rate law is:
![r=d[A]/dt=-k[A]](https://tex.z-dn.net/?f=r%3Dd%5BA%5D%2Fdt%3D-k%5BA%5D)
Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:
![[A]=[A]_0e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_0e%5E%7B-kt%7D)
You know [A]₀, k, and t, thus you can calculate [A].
![[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}](https://tex.z-dn.net/?f=%5BA%5D%3D0.548M%5Ctimes%20e%5E%7B-3.6%5Ccdot%2010%5E%7B-4%7D%2Fs%5Ctimes99.2s%7D)
![[A]=0.529M](https://tex.z-dn.net/?f=%5BA%5D%3D0.529M)
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong
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When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
Learn more about balancing equation here:
brainly.com/question/26227625
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