The rate constants for a first order decay reaction are found to be:
k287.3∘C=1 x 10−4 s−1
k327.4∘C=19.2 x 10−4 s−1
(a) Determine the value, in s-1 of k250∘C
(b) How long will it take, at 350 ∘C, for the reactant to decay to 1% of its original concentration? Express your answer in seconds.
a)4.27*10^-6
b)536.7
aja MITx FINAL EXAM
To determine the value of k250∘C, we can use the Arrhenius equation, which is given by:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
Given that we have the rate constants at 287.3∘C and 327.4∘C, we need to convert these temperatures to Kelvin.
T₁ = 273.15 + 287.3 = 560.45 K
T₂ = 273.15 + 327.4 = 600.55 K
We can then set up a ratio between the two rate constants:
k₁/k₂ = exp((Ea/R) * (1/T₂ - 1/T₁))
Plugging in the values:
1 x 10^(-4) s^(-1) / (19.2 x 10^(-4) s^(-1)) = exp((Ea/8.314) * (1/600.55 - 1/560.45))
Simplifying the equation:
0.052083 = exp(0.121602 * (0.001778 - 0.001784))
0.052083 = exp(-0.00006732 * (-0.000006))
0.052083 = exp(4.0369e-13)
To solve for k250∘C, we need to use the same equation:
k = Ae^(-Ea/RT)
We can rearrange the equation to solve for A:
A = k * e^(Ea/RT)
Plugging in the value of k250∘C (which is what we're trying to find), R, and the temperature T (250∘C converted to Kelvin):
A = k * e^(Ea / (8.314 * 523.15))
A = 1 x 10^(-4) s^(-1) * exp(Ea / (8.314 * 523.15))
To find the value of A, we can use the calculated value of 0.052083:
A = 0.052083 * exp(Ea / (8.314 * 523.15))
Now we can solve for k250∘C by plugging in the value of A and again using the Arrhenius equation:
k250∘C = A * e^(-Ea / (8.314 * 523.15))
Now, moving on to part (b):
To determine the time it takes for the reactant to decay to 1% of its original concentration at 350 ∘C, we need to use the first-order decay equation:
[A] = [A₀] * e^(-kt)
where [A] is the concentration at time t, [A₀] is the initial concentration, k is the rate constant, and t is the time.
We want to solve for t when [A] = 0.01[A₀], so:
0.01[A₀] = [A₀] * e^(-kt)
0.01 = e^(-kt)
Taking the natural logarithm of both sides:
ln(0.01) = -kt
Solving for t:
t = -ln(0.01) / k
Plugging in the given value of k (19.2 x 10^(-4) s^(-1)):
t = -ln(0.01) / (19.2 x 10^(-4) s^(-1))
Now we can calculate the time in seconds.