Chemical rate equations The reaction of certain chemical compounds can be modeled using a differential equation of the form y′ ( t ) = –ky n ( t ), where y ( t ) is the concentration of the compound for t ≥ 0, k > 0 is a constant that determines the speed of the reaction, and n is a positive integer called the order of the reaction. Assume that the initial concentration of the compound is y (0) = y 0 > 0. a. Consider a first-order reaction ( n = 1) and show that the solution of the initial value problem is y ( t ) = y 0 e – kt . b. Consider a second-order reaction ( n = 2) and show that the solution of the initial value problem is y ( t ) = y 0 y 0 k t + 1 . c. Let y 0 = 1 and k = 0.1. Graph the first-order and second-order solutions found in parts (a) and (b). Compare the two reactions.
Chemical rate equations The reaction of certain chemical compounds can be modeled using a differential equation of the form y′ ( t ) = –ky n ( t ), where y ( t ) is the concentration of the compound for t ≥ 0, k > 0 is a constant that determines the speed of the reaction, and n is a positive integer called the order of the reaction. Assume that the initial concentration of the compound is y (0) = y 0 > 0. a. Consider a first-order reaction ( n = 1) and show that the solution of the initial value problem is y ( t ) = y 0 e – kt . b. Consider a second-order reaction ( n = 2) and show that the solution of the initial value problem is y ( t ) = y 0 y 0 k t + 1 . c. Let y 0 = 1 and k = 0.1. Graph the first-order and second-order solutions found in parts (a) and (b). Compare the two reactions.
Chemical rate equations The reaction of certain chemical compounds can be modeled using a differential equation of the form y′(t) = –kyn(t), where y(t) is the concentration of the compound for t ≥ 0, k > 0 is a constant that determines the speed of the reaction, and n is a positive integer called the order of the reaction. Assume that the initial concentration of the compound is y(0) = y0 > 0.
a. Consider a first-order reaction (n = 1) and show that the solution of the initial value problem is y(t) = y0e–kt.
b. Consider a second-order reaction (n = 2) and show that the solution of the initial value problem is
y
(
t
)
=
y
0
y
0
k
t
+
1
.
c. Let y0 = 1 and k = 0.1. Graph the first-order and second-order solutions found in parts (a) and (b). Compare the two reactions.
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
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