The rate constant for the 1st order decomposition of \(H_2O_2\) is given by \(log k = 15 - \frac{1.2 \times 10^2}{T}\). \(E_a\) would be:
Answer & explanation
Correct answer: option 2
The correct answer is option 2.\(2.3\, \ kJ\).
To find the activation energy \(E_a\) from the given equation for the rate constant, we can use the Arrhenius equation:
\(k = A e^{-\frac{E_a}{RT}}\)
where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.
Given the expression for the rate constant:
\(\log k = 15 - \frac{1.2 \times 10^2}{T}\)
we can convert this to a more useful form by recognizing that:
\(\log k = \log A - \frac{E_a}{2.303RT}\)
where \(\log k\) can be expressed as:
\(\log k = 15 - \frac{1.2 \times 10^2}{T}\)
To match the form \(\log k = \log A - \frac{E_a}{2.303R}\), we can compare the coefficients:
\(-\frac{E_a}{2.303R} = -1.2 \times 10^2\)
So, we have:
\(\frac{E_a}{2.303R} = 1.2 \times 10^2\)
\(⇒ E_a = 1.2 \times 10^2 \times 2.303 \times R\)
The value of \(R\) is \(8.314 \, \text{J} \, \text{mol}^{-1} \text{K}^{-1}\).
\(E_a = 1.2 \times 10^2 \times 2.303 \times 8.314 \, \text{J} \, \text{mol}^{-1}\)
\(⇒ E_a \approx 2297.65\, \ J\)
\(⇒ E_a \approx 2.3\, \ kJ\)