For a first order decomposition of a certain reaction, rate constant is given by the equation \(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\)) Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.
Show Hint
When you see a rate constant equation in the form \(\log k = C - B/T\), immediately identify the slope term `B` with \(E_a/(2.303R)\). This allows you to quickly set up the equation \(E_a = B \times 2.303 \times R\) to find the activation energy.
Step 1: Understanding the Concept:
The given equation relates the rate constant (\(k\)) of a reaction to temperature (\(T\)). This is a form of the Arrhenius equation, which quantitatively describes the effect of temperature on reaction rates. We can find the activation energy (\(E_a\)) by comparing the given equation to the standard logarithmic form of the Arrhenius equation. Step 2: Key Formula or Approach:
The Arrhenius equation is \(k = A e^{-E_a/RT}\).
Taking the base-10 logarithm of both sides gives:
\[ \log_{10} k = \log_{10} A - \frac{E_a}{2.303RT} \]
We will compare this standard form with the given equation:
\[ \log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T} \]
Step 3: Detailed Explanation:
By comparing the two equations, we can equate the temperature-dependent terms:
\[ \frac{E_a}{2.303RT} = \frac{1 \times 10^4 K}{T} \]
The temperature \(T\) cancels from both sides of the equation:
\[ \frac{E_a}{2.303R} = 1 \times 10^4 K \]
Now, we can solve for the activation energy, \(E_a\):
\[ E_a = (1 \times 10^4) \times 2.303 \times R \]
Substitute the value of R = 8.314 J K⁻¹ mol⁻¹:
\[ E_a = 10000 \times 2.303 \times 8.314 \text{ J mol}^{-1} \]
\[ E_a = 191471.42 \text{ J mol}^{-1} \]
The question asks for the answer in kJ mol⁻¹, so we divide by 1000:
\[ E_a = \frac{191471.42}{1000} \text{ kJ mol}^{-1} = 191.47 \text{ kJ mol}^{-1} \]
This value is closest to 191.1 kJ mol⁻¹. Step 4: Final Answer:
The activation energy of the reaction is approximately 191.1 kJ mol⁻¹. Therefore, option (D) is correct.
