Answer:
the question please? illl help
Answer:
Thus, the order of the reaction is 2.
The rate constant of the graph which is :- 2.00 M⁻¹s⁻¹
Explanation:
The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.
The concentration vs time graph of zero order reactions is linear with negative slope.
The concentration vs time graph for a first order reactions is a exponential curve. For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.
The concentration vs time graph for a second order reaction is a hyberbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.
Considering the question,
A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M⁻¹s⁻¹.
<u>Thus, the order of the reaction is 2.</u>
<u>Also, slope is the rate constant of the graph which is :- 2.00 M⁻¹s⁻¹</u>
Answer:
Explanation:
![[A]_0](https://tex.z-dn.net/?f=%5BA%5D_0)
= Initial concentration = 1.28 M
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= Final concentration = ![0.17[A]_0](https://tex.z-dn.net/?f=0.17%5BA%5D_0)
k = Rate constant = 0.0632 s
t = Time taken
For first order reaction we have the relation
![kt=\ln\dfrac{[A]_0}{[A]}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{[A]}}{k}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{0.17[A]_0}}{0.0632}\\\Rightarrow t=28.037\ \text{s}](https://tex.z-dn.net/?f=kt%3D%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B%5BA%5D%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B%5BA%5D%7D%7D%7Bk%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B0.17%5BA%5D_0%7D%7D%7B0.0632%7D%5C%5C%5CRightarrow%20t%3D28.037%5C%20%5Ctext%7Bs%7D)
Time taken to reach the required concentration would be .
