The rate constant of a first-order reaction is \(4.60 \times\) \(10^{-4} \mathrm{~s}^{-1}\) at \(350^{\circ} \mathrm{C}\). If the activation energy is 104 \(\mathrm{kJ} / \mathrm{mol}\), calculate the temperature at which its rate constant is \(8.80 \times 10^{-4} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
The new temperature at which the rate constant of the reaction is \(8.80 × 10^{-4} s^{-1}\) is calculated by following the steps shown above using Arrhenius equation.

Step by step solution

01

Write down the known values

We are given three values: the first rate constant \(k_1 = 4.60 × 10^{-4} s^{-1}\), the activation energy \(E_a = 104 kJ/mol\) or \(104000 J/mol\) and the first temperature \(T_1 = 350°C = 350 + 273.15 = 623.15 K\), and we are trying to solve for the new temperature \(T_2\) at which the rate constant is \(k_2 = 8.80 × 10^{-4} s^{-1}\). The gas constant \(R = 8.314 JK^{-1}mol^{-1}\).
02

Solve for \(lnA\)

From Arrhenius' equation in logarithmic form, we can solve for \(lnA\) using the known values of \(k_1\), \(T_1\), \(R\) and \(E_a\). We have \(lnA = lnk_1 + \frac{E_{a}}{R}(\frac{1}{T_1})\). This gives us the pre-exponential factor, \(A\).
03

Substitute and Solve for \(T_2\)

Substitute the value of \(lnA\) into the logarithmic form of the Arrhenius equation for \(T_2\). Rearranging the equation gives us \(T_2 = \frac{1}{\frac{E_a}{R} - \frac{lnk_2}{lnA}}\) (after simplification). Substitute the respective values and solve for \(T_2\).

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