JEE Challenger
More from Chemical Kinetics
ChemMedAdv 2009 • Paper 2

Calculation of Pre-Exponential Factor and Activation Energy

For a first order reaction A→P\text{A} \rightarrow \text{P}, the temperature (T\text{T}) dependent rate constant (kk) was found to follow the equation log⁡k=−(2000)1T+6.0\log k = -(2000)\frac{1}{\text{T}} + 6.0. The pre-exponential factor A\text{A} and the activation energy Ea\text{E}_\text{a}, respectively, are

Options

A

1.0×106 s−11.0 \times 10^6\text{ s}^{-1} and 9.2 kJ mol−19.2\text{ kJ mol}^{-1}

B

6.0 s−16.0\text{ s}^{-1} and 16.6 kJ mol−116.6\text{ kJ mol}^{-1}

C

1.0×106 s−11.0 \times 10^6\text{ s}^{-1} and 16.6 kJ mol−116.6\text{ kJ mol}^{-1}

D

1.0×106 s−11.0 \times 10^6\text{ s}^{-1} and 38.3 kJ mol−138.3\text{ kJ mol}^{-1}

Correct

Topics & Concepts

Chemical KineticsArrhenius EquationRate of Reaction

Step-by-Step Solution

According to the Arrhenius equation, the temperature dependence of the rate constant kk is given by:
k=Ae−EaRTk = A e^{-\frac{E_{\text{a}}}{RT}}

Taking the base-10 logarithm on both sides:
log⁡10k=log⁡10A−Ea2.303RT\log_{10} k = \log_{10} A - \frac{E_{\text{a}}}{2.303 R T}

Rearranging in the form of a linear equation:
log⁡k=−(Ea2.303R)1T+log⁡A\log k = -\left(\frac{E_{\text{a}}}{2.303 R}\right) \frac{1}{T} + \log A

The given experimental equation is:
log⁡k=−(2000)1T+6.0\log k = -(2000)\frac{1}{T} + 6.0

Comparing the two equations:

  1. Pre-exponential factor (AA):
    log⁡10A=6.0\log_{10} A = 6.0
    A=106.0=1.0×106 s−1A = 10^{6.0} = 1.0 \times 10^6\text{ s}^{-1}
    (Since the reaction is first-order, the unit of both kk and AA is s−1\text{s}^{-1}.)

  2. Activation energy (EaE_{\text{a}}):
    Ea2.303R=2000 K\frac{E_{\text{a}}}{2.303 R} = 2000\text{ K}

Using the universal gas constant R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}:
Ea=2000×2.303×8.314 J mol−1E_{\text{a}} = 2000 \times 2.303 \times 8.314\text{ J mol}^{-1}
Ea≈38294 J mol−1≈38.3 kJ mol−1E_{\text{a}} \approx 38294\text{ J mol}^{-1} \approx 38.3\text{ kJ mol}^{-1}

Thus, the pre-exponential factor is 1.0×106 s−11.0 \times 10^6\text{ s}^{-1} and the activation energy is 38.3 kJ mol−138.3\text{ kJ mol}^{-1}.

Therefore, the correct option is D.

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