MasteringChemistry: Chapter 14 Homework-Google Chrome Secure l https://session.m

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MasteringChemistry: Chapter 14 Homework-Google Chrome Secure l https://session.masteringchemistry.com/myct/itemview.assig Most of the time, the rate of a reaction depends on the concentration of the reactant. In the case of second-order ime, s versus time is always a straight line with a slope k and a y intercept reactions, the rate is proportional to the square of the concentration of the reactant. Alo 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. Part B Consider the second-order reaction 2HI(g)H2(g) + 12(g)Use the simulation to find the initial concentration 1 0 and the rate constant k for the reaction. What will be the concentration of HI after t = 3.47s 1010 s (HIt) for a reaction starting under the condition in the simulation? Express your answer in moles per liters to three significant figures. View Available Hint(s) 64-1010d mol Submit 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. Part C This question will be shown after you complete previous question(s) Part D This question will be shown after you complete previous question(s) 9:16 PM O Type here to search 2/13/2018 3

Explanation / Answer

Given data,

k = 6.4 x 10-9 l/mol.s

and initial rate rA = 1.6 x 10-7 mol/l.s

rA = kC02

1.6 x 10-7 mol/l.s = 6.4 x 10-9 l/mol.s x C02

C02 = 25 (mol/l)2

C0 = 5 mol/l = initial concentration

So integrating the equation

-dC/dt = kC2

we get,

1/C - 1/C0 = kt

putting all the values we get:

1/C - 1/5 = 6.4 x 10-9 l/mol.s x 3.47×1010 s

1/C = 222.28 l/mol

C = 0.00449mol/l

or

[HI]t=4.49 x 10-3 M is final concentration after given time

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