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.
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)
A. Expressing y in terms of t and y0:
To solve the differential equation dy/dt = −0.05y^2, we can separate the variables:
1/(y^2) * dy = -0.05 dt
Integrating both sides:
∫(1/(y^2)) * dy = ∫(-0.05) * dt
Using the power rule:
-1/y = -0.05t + C
Where C is the constant of integration.
Solving for y:
y = -1/(-0.05t + C)
B. Finding the initial concentration of NO2:
Given that the concentration of NO2 is twice as high at t = 20 seconds compared to t = 100 seconds, we can use this information to determine the initial concentration.
Using the equation y = -1/(-0.05t + C), we can substitute t = 20 and t = 100 to form two equations:
y1 = -1/(-0.05(20) + C) .....(1)
y2 = -1/(-0.05(100) + C) .....(2)
It is given that y1 = 2y2, so we can substitute this into equation (1) to solve for C:
2y2 = -1/(-0.05(20) + C)
Rearranging the equation:
2y2(-0.05(20) + C) = -1
Simplifying:
-2(0.05(20) - C) = 1/y2
Expanding and rearranging:
-0.2 + 2C = 1/y2
Simplifying further:
2C = 1/y2 + 0.2
Substituting y2 = -1/(-0.05(100) + C):
2C = 1/(-1/(-0.05(100) + C)) + 0.2
Simplifying:
2C = (100C - 100)(-20) + 0.2
Expanding and arranging terms:
2C = -2000C + 200000 + 0.2
Combining like terms:
2002C = 200000.2
Finally, solving for C:
C = 200000.2/2002
Therefore, the exact initial concentration of NO2 is y0 = C.
A. To find the expression for y in terms of t and the initial concentration y0, we need to solve the given differential equation dy/dt = -0.05y^2.
Separating variables, we have:
dy/y^2 = -0.05dt
Integrating both sides, we get:
∫(dy/y^2) = ∫(-0.05dt)
Integrating the left side:
-1/y = -0.05t + C1,
where C1 is the constant of integration.
Now, multiplying both sides by -1, we get:
y = 1/(-0.05t + C1),
Let's call (-0.05t + C1) as C2 (new arbitrary constant) for simplicity:
y = 1/C2.
To determine C2, we can use the initial condition y(0) = y0:
y(0) = 1/C2 = y0,
C2 = 1/y0.
Therefore, the expression for y in terms of t and the initial concentration y0 is:
y(t) = 1/(1/y0) = y0.
B. We are given that at t = 20 seconds, the concentration of NO2 is twice as high as it is at t = 100 seconds. Mathematically, this can be written as:
y(20) = 2y(100).
Using the expression for y(t) found in part A, we can substitute the values:
y0 = 2y0.
Simplifying, we find:
2y0 = y0,
Since this equation is true for all values of y0, we can conclude that the initial concentration of NO2 can be any value.
Hence, the exact initial concentration of NO2 cannot be determined solely based on the given information.
To find the concentration of NO2, y(t), in terms of time, t, and the initial concentration, y0, we need to solve the differential equation dy/dt = -0.05y^2. This is a separable differential equation.
Here's the step-by-step process:
Step 1: Separation of variables
Let's rewrite the equation as:
dy / y^2 = -0.05dt
Step 2: Integration
Now, integrate both sides of the equation:
∫ (1/y^2) dy = ∫ (-0.05) dt
-1/y = -0.05t + C
Step 3: Solve for y
To find the constant, C, we need to use the initial condition, y(t=0) = y0:
-1/y0 = -0.05(0) + C
-1/y0 = C
Substituting the value of C back into the equation:
-1/y = -0.05t - (1/y0)
Step 4: Rearrange the equation for y
Multiply both sides by y and isolate y on one side:
-1 = -0.05yt - y/y0
Now, add y/y0 to both sides:
-1 + y/y0 = -0.05yt
Finally, divide both sides by -0.05t:
(y/y0 - 1)/(-0.05t) = 1
Multiplying both sides by -0.05t:
y/y0 - 1 = -0.05t
And adding 1 to both sides:
y/y0 = -0.05t + 1
Finally, multiplying both sides by y0:
y = (-0.05t + 1) * y0
So the expression for y(t) in terms of t and the initial concentration y0 is:
y(t) = (-0.05t + 1) * y0
Next, we need to find the exact initial concentration of NO2, y0, assuming that the concentration of NO2 is twice as high at t = 20 seconds as it is at t = 100 seconds.
Let's substitute the values into the equation for y(t):
y(20) = (-0.05 * 20 + 1) * y0 = (1 - 1) * y0 = 0
And y(100) = (-0.05 * 100 + 1) * y0 = (1 - 5) * y0 = -4y0
Given that y(20) = 2 * y(100), we can set up the equation:
0 = 2 * (-4y0)
Simplifying:
0 = -8y0
Since the concentration cannot be negative, we can conclude that y0 is equal to 0.
Therefore, the exact initial concentration of NO2, y0, is 0 moles per liter.