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Q3.27:
The rate constant for the first order decomposition of (H_{2}O_{2}) is given by the following equation: (\log k = 14.34 – 1.25 \times 10^{4}K/T) Calculate (E_{a}) for this reaction and at what temperature will its half-period be 256 minutes?

Solution :

Step 1: Identifying the Arrhenius Equation Constants

The Arrhenius equation in logarithmic form is given by:

$\ln k = \ln A - \frac{E_{a}}{RT}$

Converting to base-10 logarithm ($\log_{10}$):

$\log k = \log A - \frac{E_{a}}{2.303RT}$ [Per the logarithmic identity $\ln x = 2.303 \log x$]

Given equation: $\log k = 14.34 - \frac{1.25 \times 10^{4}}{T}$

By comparing the coefficients of $1/T$, we establish:

$\frac{E_{a}}{2.303R} = 1.25 \times 10^{4} \, K$

Step 2: Calculating Activation Energy ($E_{a}$)

Using the gas constant $R = 8.314 \, J \, K^{-1} \, mol^{-1}$:

$E_{a} = 1.25 \times 10^{4} \, K \times 2.303 \times 8.314 \, J \, K^{-1} \, mol^{-1}$

$E_{a} = 1.25 \times 10^{4} \times 19.147 \, J/mol$

$E_{a} = 239337.5 \, J/mol$

$E_{a} = 239.34 \, kJ/mol$

Step 3: Determining the Rate Constant ($k$) for $t_{1/2} = 256 \, minutes$

For a first-order reaction, the relationship between half-life and rate constant is:

$k = \frac{0.693}{t_{1/2}}$ [Per the derivation of integrated rate laws for first-order kinetics]

$k = \frac{0.693}{256 \, min} = 0.002707 \, min^{-1}$

To use this in the logarithmic equation, we convert to seconds (standard scientific practice):

$k = \frac{0.002707}{60} \, s^{-1} = 4.51 \times 10^{-5} \, s^{-1}$

Step 4: Solving for Temperature ($T$)

Substitute $k$ into the given equation:

$\log(4.51 \times 10^{-5}) = 14.34 - \frac{1.25 \times 10^{4}}{T}$

$\log(4.51) + \log(10^{-5}) = 14.34 - \frac{1.25 \times 10^{4}}{T}$

$0.654 - 5 = 14.34 - \frac{1.25 \times 10^{4}}{T}$

$-4.346 = 14.34 - \frac{1.25 \times 10^{4}}{T}$

$\frac{1.25 \times 10^{4}}{T} = 14.34 + 4.346 = 18.686$

$T = \frac{1.25 \times 10^{4}}{18.686} \approx 668.95 \, K$

Inverse Temperature (1/T) log k Slope = -Ea / 2.303R

Final Solution: Activation Energy ($E_{a}$) = $239.34 \, kJ/mol$; Temperature ($T$) $\approx 668.95 \, K$


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