PHY1610 Assignment 5: Consider a chemical reaction between the three substances…solution




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Assignment 5:
Consider a chemical reaction between the three substances X, Y, Z in solution (well-mixed).

2\textrm X + Y \rightleftharpoons 2 Z

Let us denote the concentrations of X, Y, and Z by x, y, and z.

Let the rate of the forward reaction be k1, and that of the backward reaction be k2.

According to chemical kinetics, in equilibrium, the concentrations will satisfy


This single equation is not enough to determine the three concentrations.

Additional information comes from the chemical reaction itself, which shows that x-2y is conserved in the reaction, and so is x+z. In other words,

x-2y = c_1

x+z = c_2

Where c1 and c2 are set by the initial concentrations of the substances, denoted by x0, y0 and z0. I.e. c_1=x_0-2y_0 and c_2=x_0+z_0.

Part 1

We are after the equilibrium values of x, y and z given the following parameters and initial concentrations:

k_1=1, k_2=0.7, x_0 = 0.5, y_0 = 1, z_0 = 0

Solve for the equilibrium concentrations in two ways

Directly using the above three equations, using a three-dimensional root finding routine from the GSL.
Using that one can combine the above three equations into a single one for x:

k_1x^3-(c_1k_1+2k_2)x^2+4c_2k_2 x-2k_2 c_2^2 = 0

Use one of the polynomial root finding routines from the GSL for this second part.

Part 2

We are after the time it takes to reach this equilibrium. For that, we need the rate equations of the above chemical reaction, which are a set of three ODEs:

\frac{dx}{dt} = 2k_2z^2 – 2k_1x^2y

\frac{dy}{dt} = k_2z^2 – k_1x^2y

\frac{dz}{dt} = 2k_1x^2y – 2k_2z^2

Take the same initial conditions as in part 2, and solve these ODEs using the GSL’s odeiv2 runge-kutta method. Solve it from time 0 to time t, such that you keep increasing t until the concentrations vary less than 1%. What is the time t?

Compare the results for the concentrations with those found in part 1.