The question is long but I have done everything and I need only part B Most of t

Question

The question is long but I have done everything and I need only part B

Most of the time, the rate of a reaction depends on the concentration of the reactant. In the case of second-order reactions, the rate is proportional to the square of the concentration of the reactant.

Select the image to explore the simulation, which will help you to understand how second-order reactions are identified by the nature of their plots. You can also observe the rate law for different reactions.

In the simulation, you can select one of the three different kinds of plots. You may use the Start , Stop, and Reset buttons to observe the corresponding changes in the plot for different kinds of reactions. You can also select six different reactions using the drop-down menu and observe three different types of plots for each reaction.

Relating plots to the order of a reaction

Consider the following reaction:

Aproducts

The plot of [A] versus t is linear for the zero-order reaction, the plot of ln[A] versus t is linear for the first-order reaction, and the plot of 1[A] versus t is linear for the second-order reaction. [A] represents the concentration of the reactant A.

The linearity of each graph can be used to identify the order of a reaction.

Part A

There are six different reactions you can access in the simulation using the drop-down menu. Which of the following are second-order reactions?

Check all that apply.

Correct

Characteristics of second-order reactions

For a second-order reaction, [A]products, the rate of the reaction is given as rate= k[A]2, where k is the rate constant and [A] is the concentration of reactant A. The integrated rate law for second-order reactions is 1[A]t=kt+1[A]0, where [A]t is the concentration of reactant A at time t, k is the rate constant, and [A]0 is the initial concentration of reactant A. This equation is of the type y=mx+b. Therefore, the plot of 1[A]t versus time is always a straight line with a slope k and a y intercept 1[A]0.

Part B

Consider the second-order reaction:

2HI(g)H2(g)+I2(g)

Use the simulation to find the initial concentration [HI]0 and the rate constant k for the reaction. What will be the concentration of HI after t = 5.19×1010 s ([HI]t) for a reaction starting under the condition in the simulation?

Express your answer in moles per liters to three significant figures.

I need part B

Explanation / Answer

Given that

the initial concentration [HI]0 =1.6*10-3 mol/l-s

the rate constant k=6.4*10^-9 l / mol-s

integrated rate law for 2nd order reaction:
1/[At] = +kt + 1/[Ao]

you have..
[At] = ?

[Ao] = 1.6*10-3 M
t =5.19*10^10 s

1/[At] = +6.4*10^-9 l / mol-s*5.19*10^10 s + 1/1.6*10-3 mol/l

1/[At] =332.16 l / mol+ 625 l / mol

[At] =1.045 *10^-3 mol/L

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