The rate constant for the first order decomposition of H₂O₂ is given by the following equation: log k = 14.34 - 1.25 × 10⁴ K/T Calculate Eₐ for this reaction and at what temperature will its half-period be 256 minutes?

Activation energy Eₐ is 239.3 kJ mol⁻¹ and the temperature T is 669 K. — To find the activation energy, we compare the given equation with the logarithmic form of the Arrhenius equation provided in the context. The slope term allows us to solve for Eₐ using the gas constant R. For the second part, the rate constant k is derived from the given half-life using the first-order kinetics formula, and this value is substituted back into the logarithmic equation to solve for temperature T.

The rate constant for the first order decomposition of H₂O₂ is given…

Question 34 of 37advanced🔧 ApplyNumerical3 marks

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?

Correct Answer

Activation energy $E_a$ is 239.3 kJ mol $^{-1}$ and the temperature $T$ is 669 K.

Exercise: EXERCISES | Q: 3.27 | (Page 28)

For More Understanding

Explanation

To find the activation energy, we compare the given equation with the logarithmic form of the Arrhenius equation provided in the context. The slope term allows us to solve for $E_a$ using the gas constant $R$ . For the second part, the rate constant $k$ is derived from the given half-life using the first-order kinetics formula, and this value is substituted back into the logarithmic equation to solve for temperature $T$ .

Solution Steps

Step 1: Calculate Activation Energy ( $E_a$ ) The given equation is $\log k = 14.34 - \frac{1.25 \times 10^4 K}{T}$ . Comparing this with the Arrhenius equation form $\log k = \log A - \frac{E_a}{2.303RT}$ , we equate the slope terms: $\frac{E_a}{2.303R} = 1.25 \times 10^4 K$ . Using $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ (from Example 3.9 context),
$E_a = 1.25 \times 10^4 \times 2.303 \times 8.314$
$E_a = 239339.77 \text{ J mol}^{-1} \approx 239.3 \text{ kJ mol}^{-1}$ .
Step 2: Calculate Rate Constant ( $k$ ) Given half-period ( $t_{1/2}$ ) = 256 minutes = $256 \times 60 = 15360$ s. For a first-order reaction, $k = \frac{0.693}{t_{1/2}}$ . $k = \frac{0.693}{15360} = 4.51 \times 10^{-5} \text{ s}^{-1}$ .
Step 3: Calculate Temperature ( $T$ ) Substitute the value of $k$ into the given equation:
$\log(4.51 \times 10^{-5}) = 14.34 - \frac{1.25 \times 10^4}{T}$ $-4.345 = 14.34 - \frac{12500}{T}$ $\frac{12500}{T} = 14.34 + 4.345 = 18.685$
$T = \frac{12500}{18.685} = 668.9 \text{ K} \approx 669 \text{ K}$ .

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