The relationship between the rate of a certain chemical reaction and temperature under certain circumstances is given byR(T)=0.1(−0.05T^3+4T^2+120)grams/sec, where Ris the rate of reaction and Tis the temperature (in °C).
a)Find thetemperature T at which the reaction rate reaches its maximum?
b)What is the maximum reaction rate?
It will reach a max when R '(T) = 0
R'(T) = .1(-.15T^2 + 8T) = 0
.15t^2 - 8t = 0
t(.15t - 8) = 0
t = 0 , (at the beginning)
or
t = 8/.15 sec = 160/3 sec
find R(160/3) to get that maximum
To find the temperature at which the reaction rate reaches its maximum (question a) and the maximum reaction rate (question b), we can use calculus to find the maximum point of the function R(T).
a) To find the temperature at which the reaction rate reaches its maximum, we need to find the critical points of the function R(T). Critical points occur where the derivative of the function is equal to zero or is undefined.
1. Start by finding the derivative of R(T) with respect to T:
dR(T)/dT = 0.1((-0.05 * 3 * T^3) + (4 * 2 * T^2))
2. Set the derivative equal to zero and solve for T:
0.1((-0.15 * T^3) + (8 * T^2)) = 0
Simplifying the equation:
-0.015T^3 + 0.8T^2 = 0
Factoring out T:
T ( -0.015T^2 + 0.8 ) = 0
Setting each factor equal to zero:
T = 0 (one of the critical points)
-0.015T^2 + 0.8 = 0
3. Solve the second equation:
-0.015T^2 = -0.8
Divide both sides by -0.015:
T^2 = 53.333...
Taking the square root of both sides:
T ≈ ±7.31
Since the problem specifies temperature in °C, we can ignore the negative value and conclude that the critical temperature is approximately T = 7.31 °C.
b) To find the maximum reaction rate, substitute the critical temperature T = 7.31 °C back into the function R(T):
R(T) = 0.1(-0.05 * 7.31^3 + 4 * 7.31^2 + 120)
Using a calculator to evaluate the expression:
R(T) ≈ 12.61 grams/sec
Therefore, the maximum reaction rate is approximately 12.61 grams/sec.