Question

If for a first order reaction, the value of $A$ and $E_a$ are $4 \times 10^{13} s^{-1}$ and $98.6 kJ mol^{-1}$ respectively, then at what temperature will its half life period be 10 minutes?

• A. 330 K
• B. 300 K
• C. 330.95 K
• D. 311.15 K

Answer 1

Answer: (D)

311.15 K

Answer 2

The half-life period ($t_{1/2}$) for a first-order reaction is related to the rate constant ($k$) by the equation:

$t_{1/2} = \frac{0.693}{k}$

Given $t_{1/2} = 10$ minutes $= 10 \times 60 = 600$ seconds. So, the rate constant $k$ is:

$k = \frac{0.693}{600} = 0.001155 \, s^{-1} = 1.155 \times 10^{-3} \, s^{-1}$

The Arrhenius equation relates the rate constant ($k$), the pre-exponential factor ($A$), the activation energy ($E_a$), the gas constant ($R$), and the temperature ($T$):

$k = A e^{-E_a/RT}$

We can take the natural logarithm of both sides:

$\ln(k) = \ln(A) - \frac{E_a}{RT}$

Rearranging the equation to solve for $T$:

$\frac{E_a}{RT} = \ln(A) - \ln(k) = \ln\left(\frac{A}{k}\right)$

$T = \frac{E_a}{R \ln\left(\frac{A}{k}\right)}$

Alternatively, using base-10 logarithm:

$\log_{10}(k) = \log_{10}(A) - \frac{E_a}{2.303 RT}$

$\frac{E_a}{2.303 RT} = \log_{10}(A) - \log_{10}(k) = \log_{10}\left(\frac{A}{k}\right)$

$T = \frac{E_a}{2.303 R \log_{10}\left(\frac{A}{k}\right)}$

We are given:

$A = 4 \times 10^{13} \, s^{-1}$

$E_a = 98.6 \, kJ \, mol^{-1} = 98.6 \times 10^3 \, J \, mol^{-1}$

$R = 8.314 \, J \, K^{-1} \, mol^{-1}$

First, calculate $\frac{A}{k}$:

$\frac{A}{k} = \frac{4 \times 10^{13} \, s^{-1}}{1.155 \times 10^{-3} \, s^{-1}} = \frac{4}{1.155} \times 10^{13 - (-3)} = 3.4632 \times 10^{16}$

Now, calculate $\log_{10}\left(\frac{A}{k}\right)$:

$\log_{10}\left(\frac{A}{k}\right) = \log_{10}(3.4632 \times 10^{16}) = \log_{10}(3.4632) + \log_{10}(10^{16})$

$\log_{10}(3.4632) \approx 0.53945$

$\log_{10}\left(\frac{A}{k}\right) \approx 0.53945 + 16 = 16.53945$

Now, substitute the values into the formula for $T$:

$T = \frac{E_a}{2.303 R \log_{10}\left(\frac{A}{k}\right)}$

$T = \frac{98.6 \times 10^3 \, J \, mol^{-1}}{2.303 \times 8.314 \, J \, K^{-1} \, mol^{-1} \times 16.53945}$

$T = \frac{98600}{19.147142 \times 16.53945}$

$T = \frac{98600}{316.659}$

$T \approx 311.34 \, K$

Comparing this value with the given options:

A 330 K B 300 K C 330.95 K D 311.15 K

The calculated value 311.34 K is closest to option D (311.15 K). The slight difference might be due to rounding in the value of 0.693 or the value of R used. Using the value 0.6931 for $\ln(2)$ gives $k = 0.6931/600 \approx 0.001155166$.

$\log_{10}(A/k) = \log_{10}(4 \times 10^{13} / 0.001155166) = \log_{10}(34626886 \times 10^9) = \log_{10}(3.4626886 \times 10^{16}) \approx 16.5394$.

$T = \frac{98600}{19.147142 \times 16.5394} = \frac{98600}{316.65} \approx 311.34 K$.

Using $k = 0.693/600$ and $\log_{10}(A/k) \approx 16.5395$ as calculated before, the value is 311.34 K. Option D is the closest value.

The final answer is 311.15 K.