Question

4. At high temperatures, nitrogen dioxide, NO2, decomposes into NO and O2. If y(t) is the concentration of NO2 (in moles per liter), then at 600 degrees K, y(t) changes according to the
reaction law dy/dt = −. 05y2 for time t in seconds.
A. Express y in terms of t and the initial concentration y0.
B. Assuming that the concentration of NO2 is twice as high at t = 20 seconds as it is at 100 seconds, find the exact initial concentration of the NO2.

Answer 1

A.
dy/dt = - 0.05∙y²
That is a separable first order differential equation: Separation of variables leads to:
- (1/y²) dx = 0.05 dt
Hence,
∫ - (1/y²) dx = ∫ 0.05 dt
=>
(1/y) = 0.05∙t + C
Let y(0) = y₀
(1/y₀) = 0.05∙0 + C
C = (1/y₀)

(1/y) = 0.05∙t + (1/y₀)
y(t) = 1/(0.05∙t + (1/y₀)) = y₀/(y₀∙0.05∙t + 1)


B.
y(20)/y(100) = 2
[y₀/(y₀∙0.05∙20 + 1)] / [y₀/(y₀∙0.05∙100 + 1)] = 2
(5∙y₀+ 1)/(y₀ + 1) = 2
5∙y₀+ 1 = 2∙y₀ + 2
y₀ = (1/3)