A second order chemical reaction involves the interaction (collision) of one molecule of a substance p with one molecule of a substance q to produce one molecule of a new substance x; this is denoted by p + q → x. suppose that p and q, where p = q, are the initial concentrations of p and q, respectively, and let x(t) be the concentration of x at time t. then p − x(t) and q − x(t) are the concentrations of p and q at time t, and the rate at which the reaction occurs is given by the equation dx/dt = α(p − x)(q − x), (i) where α is a positive constant. (a) if x(0) = 0, determine the limiting value of x(t) as t → ∞ without solving the differential equation. then solve the initial value problem and find x(t) for any t.

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22-09-2017
Physics

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A second order chemical reaction involves the interaction (collision) of one molecule of a substance p with one molecule of a substance q to produce one molecule of a new substance x; this is denoted by p + q → x. suppose that p and q, where p = q, are the initial concentrations of p and q, respectively, and let x(t) be the concentration of x at time t. then p − x(t) and q − x(t) are the concentrations of p and q at time t, and the rate at which the reaction occurs is given by the equation dx/dt = α(p − x)(q − x), (i) where α is a positive constant. (a) if x(0) = 0, determine the limiting value of x(t) as t → ∞ without solving the differential equation. then solve the initial value problem and find x(t) for any t.

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shinmin shinmin · Answer 1 · 2017-10-05T16:51:21+02:00

From (i) it’s indistinct that there are two equilibrium solutions x = p and x = q. Since p 6= q we might as well accept p < q (or else we just switch the roles of p and q). By examining the derivatives of f(x) = α(p − x)(q − x) at p and q we see that f 0 (p) = −α(q − p) < 0 and f 0 (q) > 0. Therefore x = p is an asymptotically stable solution and we expect x → p as t → ∞

(see attached file for the solution) the answer would be if we use the initial condition we get C = p/q.