Download the App!
Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.
Question
Answered step-by-step
In contrast to the situation of Exercise 40, experiments show that the reaction $ H_2 + Br_2 \to 2HBr $ satisfies the rate law$ \frac {d[HBr]}{dt} = k[H_2][Br_2]^{1/2} $and so for this reaction the differential equation becomes $ \frac {dx}{dt} = k(a - x)(b - x)^{1/2} $where $ x = [HBr] $ and $ a $ and $ b $ are the initial concentrations of hydrogen and bromine.(a) Find $ x $ as a function of $ t $ in the case where $ a = b. $ Use the fact that $ x(0) = 0. $(b) If $ a > b, $ find $ t $ as a function of $ x. $ [Hint: In performing the integration, make the substitution $ u = \sqrt {b - x}. $ ]
Video Answer
Solved by verified expert
This problem has been solved!
Try Numerade free for 7 days
Like
Report
Official textbook answer
Video by Subham Jyoti Mishra
Numerade Educator
This textbook answer is only visible when subscribed! Please subscribe to view the answer
Calculus 2 / BC
Chapter 9
Differential Equations
Section 3
Separable Equations
Baylor University
University of Michigan - Ann Arbor
Idaho State University
Boston College
Lectures
13:37
A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.
33:32
12:28
In contrast to the situati…
01:45
04:21
The gas-phase reaction bet…
03:55
The first-order integrated…
11:36
Consider the reaction:…
06:50
In this case, we have the differential equation: d x by d. T is: got a minus x into p minus x into minus 6 to the power 1 by 2, and we have the initial condition. Y of 0 is equal to 0. So in first case case, we have a is equal to the differential becomes texts. A minus 6 to the power 3 by 2 to the power. A 1 to 3 by 2 is equal to k d. T integration, both the sides- minus a minus x to the power 3 by 2- is the quarto plus 1 upon a minus x to the power minus s 3 by 2 plus 1 is equal to 84 cioactwo 2, a minus 6 minus 1 by 2 sepolte t Plus 3 or 2 by root, a minus x is equal to 50 plus c. We get root. A minus x is equal to 2 by 8 plus c square in a minus x is equal to 4. On k. T plus whole square and x is equal to a minus 4 by 8 is c, whole square is solis, we have that t is equal to 0 x is equal to 0 plus 0 is equal to minus 2. Pi square is for is square c square. Is equal to 4 pi a which in plus c, is equal to 2 pi otasite fore the equation becomes. This is a minus 4 pi k, t plus 2 by root a whole squarethis is part a of the salatin part b. We have the condition that is greater than v, so d x d t is equal to k into minus x minus 1 by 2. So this becomes d x n a minus x to p minus x is pocatintegrating in both the sides we gettisassume root. P, minus x to d over dx is equal to minus 1 by 2. V? U or you can write the x minus 2? U! D! U substituting this here! We get x, is equal to v minus whole square. So this equation becomes! First? U d? U, upon a minus p, minus whole square and multiply by you tiresias k, t s c solving this discertion that we have minus t? U upon a minus v: minus! U square is equal to 30 plus c and we have to substitute 1 more to prosy. I, by a minus b, o a minus v, so you becomes a minute to or do is equal to t to blew root a minus b. So this is becomes minus d to root minus v, o a minus p minus w square. A minus b is equal to k, t plus c it. This becomes minus root. A minus b upon a minus b and by 1 minus double square plus c, is 1 by root. A minus b integration by 1 minus double square t k, t plus c solving this we get okay. Here we have last to 1 square a minus b, so this is equal to minus oho. We left 2 here minus 2 by root a minus b into integration, 1 plus to 10 astor k, t plus c solving this for the weed minus 2 put a minus b into 10. As is u, by a minus b? U, by foot a minus b is q, t plus c and d: u is equal to 2 p minus 6 minus 2 by root n minus b into 10 in p minus x by root a minus b e o k, t plus c so in the initial Conditions of t is equal to 00, so minus 2 y root, a minus b 10, must b by a minus b, is equal to c. The final equation becomes, t is equal to 2 bites to by root angulus tenawas root, p minus x by root a minus b minus the would be by canals drawn by k or t is equal to 2 pi k root, an onus p tan inverse root by A root: a minus b minus 10 in verse, minus x, minus- and this is a solution upon.
View More Answers From This Book
Find Another Textbook
