Question

The first order rate constant for the decomposition of CaCO\(_3\) at 700 K is 6.36 \(\times\) 10\(^{-3}\) s\(^{-1}\) and activation energy is 209 kJ mol\(^{-1}\). Its rate constant (in s\(^{-1}\)) at 600 K is \(x \times 10^{-6}\). The value of x is _________. (Nearest integer)
[Given R = 8.31 J K\(^{-1}\) mol\(^{-1}\); log 6.36\(\times\)10\(^{-3}\)=\(-2.19\), 10\(^{-4.79}\)=1.62\(\times\)10\(^{-5}\)]

Hint:

The Arrhenius equation shows that the rate constant is highly dependent on temperature. A decrease in temperature will always lead to a decrease in the rate constant. If you get an answer where the rate is higher at a lower temperature, you've likely made a sign error in the \((1/T_1 - 1/T_2)\) term.

Answer 1

Correct Answer: 16

Step 1: Understanding the Question:
We are given a rate constant at one temperature and the activation energy. We need to find the rate constant at a different, lower temperature using the Arrhenius equation.
Step 2: Key Formula:
The two-point form of the Arrhenius equation is:
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] Step 3: Substitute the Given Values:
Let:
- \(T_1 = 700\) K, \(k_1 = 6.36 \times 10^{-3}\) s\(^{-1}\).
- \(T_2 = 600\) K, \(k_2 = ?\)
- \(E_a = 209\) kJ/mol = 209000 J/mol.
- \(R = 8.31\) J K\(^{-1}\) mol\(^{-1}\).
\[ \log\left(\frac{k_2}{6.36 \times 10^{-3}}\right) = \frac{209000}{2.303 \times 8.31} \left(\frac{1}{700} - \frac{1}{600}\right) \] \[ \log(k_2) - \log(6.36 \times 10^{-3}) = \frac{209000}{19.147} \left(\frac{600 - 700}{420000}\right) \] \[ \log(k_2) - (-2.19) = (10915.6) \left(\frac{-100}{420000}\right) \] \[ \log(k_2) + 2.19 = 10915.6 \times (-2.381 \times 10^{-4}) \approx -2.60 \] \[ \log(k_2) = -2.60 - 2.19 = -4.79 \] Step 4: Calculate k\(_2\):
To find \(k_2\), we take the antilog:
\[ k_2 = 10^{-4.79} \] The problem helpfully provides this value: \(10^{-4.79} = 1.62 \times 10^{-5}\).
So, \(k_2 = 1.62 \times 10^{-5}\) s\(^{-1}\).
Step 5: Find the value of x:
We are given that the rate constant at 600 K is \(x \times 10^{-6}\).
\[ 1.62 \times 10^{-5} = x \times 10^{-6} \] To make the powers of 10 equal, we can write \(1.62 \times 10^{-5}\) as \(16.2 \times 10^{-6}\).
\[ 16.2 \times 10^{-6} = x \times 10^{-6} \] \[ x = 16.2 \] The value of x to the nearest integer is 16.