Question

The relation between half reaction time

and initial concentration of

reactant for (n – 1) order reaction is ​

Answer 1

Answer:

An alternate way to determine a rate law is to monitor the concentration of reactants or products in a single trial over a period of time and compare that to what is expected mathematically for a first-, second-, or zero-order reaction.

First-Order Reactions

We have seen earlier that the rate law of a generic first-order reaction where A → B can be expressed in terms of the reactant concentration:

Rate of reaction = – $$\frac{\Delta \ [A]}{\Delta \ t}\$$ = $$\textit{k}[A]{}^{1}\$$

This form of the rate law is sometimes referred to as the differential rate law. We can perform a mathematical procedure known as an integration to transform the rate law to another useful form known as the integrated rate law:

ln $$\frac{{[A]}_t}{{[A]}_0}\ $$= -$$\textit{k}t\$$

where “ln” is the natural logarithm, [A]0 is the initial concentration of A, and [A]t is the concentration of A at another time.

The process of integration is beyond the scope of this textbook, but is covered in most calculus textbooks and courses. The most useful aspect of the integrated rate law is that it can be rearranged to have the general form of a straight line (y = mx + b).

ln $$\ {[A]}_t\$$= -$$\textit{k}t + ln {[A]}_0\$$

(y = mx + b)

Therefore, if we were to graph the natural logarithm of the concentration of a reactant (ln) versus time, a reaction that has a first-order rate law will yield a straight line, while a reaction with any other order will not yield a straight line (Figure 17.7 “Concentration vs. Time, First-Order Reaction”). The slope of the straight line corresponds to the negative rate constant, –k, and the y-intercept corresponds to the natural logarithm of the initial concentration.

Figure 17.7. Concentration vs. Time, First-Order Reaction

Figure 17.4.1. Plot of natural logarithm of concentration versus time for a first-order reaction.

This graph shows the plot of the natural logarithm of concentration versus time for a first-order reaction.

Example 4

The decomposition of a pollutant in water at 15oC occurs with a rate constant of 2.39 y-1, following first-order kinetics. If a local factory spills 6,500 moles of this pollutant into a lake with a volume of 2,500 L, what will the concentration of pollutant be after two years, assuming the lake temperature remains constant at 15oC?

Solution

We are given the rate constant and time and can determine an initial concentration from the number of moles and volume given.

$$\ {[Pollutant]}_0\$$= $$\frac{{\rm 6500\ mol}}{{\rm 2500\ L}}\$$ = 2.6 M

We can substitute this data into the integrated rate law of a first-order equation and solve for the concentration after 2.0 years:

Solution

1. $$\ t{}_{1/2 }\$$ = $$\frac{0.693}{k}\$$ = $$\frac{0.693}{{\rm 4.00\ x\ }{{\rm 10}}^{-3}{\rm \ }{{\rm s}}^{-1}}\$$ = 173 s

2. A simple way to calculate this is to determine how many half-lives it will take to go from 1.00 M to 0.250 M and use the half-life calculated in part 1.

1 half-life = 0.500 M

2 half-lives = 0.250 M

Therefore, it will take 2 x 173 s = 346 s.

3. We can use the rate-constant value in the integrated rate law to determine the concentration remaining.

ln $$\frac{{[A]}_t}{{[A]}_0}\$$= -$$\textit{k}t\$$

ln $$\frac{{[A]}_t}{1.0\ M}\$$= -$$\ (4.00 x 10{}^{-3} s{}^{-1})(500 s)\$$

ln $$\frac{{[A]}_t}{1.0\ M}\$$ = -2

$$\frac{{[A]}_t}{1.0\ M}\$$ = $$\ e{}^{-2 }\$$= 0.135

$$\ [A]{}_{t}\$$ = 0.14 M

Key Takeaways

thank you

Explanation:

hope it's helpful you

rate me as a brainliast answer

Answer 2

hii...anna ela unnav....

naa question lo reply ivvu anna ...report this after reading....