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 \(104 \mathrm{~kJ} /\) mol, calculate the temperature at which its rate constant is \(8.80 \times 10^{-4} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
The temperature at which the rate constant is \(8.80 \times 10^{-4} s^{-1}\) is approximately \(533 °C\).

Step by step solution

01

Understand the Arrhenius equation

The Arrhenius equation is \(k=Ae^{-Ea/RT}\) where \(k\) is the rate constant, \(A\) is the frequency factor, \(Ea\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature. Taking the natural logarithm on both sides, it becomes \(ln k = ln A - Ea/RT\). We will make use of this transformed equation.
02

Plug in the given values

We know two rate constants correspond to two temperatures. We can express them as \(ln k_1 = ln A - Ea/R \times 1/T_1\) and \(ln k_2 = ln A - Ea/R \times 1/T_2\). Subtracting the two equations gives \(ln (k_2/k_1) = Ea/R \times (1/T_1 - 1/T_2)\). We have the values for \(k_1\), \(k_2\), \(R\), \(Ea\) and \(T_1\). The only unknown is \(T_2\), which is what we're asked to find.
03

Solve for T_2

Rearrange the equation to solve for \(T_2\). This gives \(T_2 = 1/ ((1/T_1) - (R/Ea) \times ln(k_2/k_1)) \). Remember to convert the activation energy to J/mol (1 kJ = 1000 J), the gas constant R to its appropriate value in J/(mol*K) (R = 8.314 J/(mol*K)), and T_1 to Kelvin (T_1= 350°C+273=623 K). Then plug in \(k_1 = 4.60 × 10^{-4} s^{-1}\), \(Ea = 104 × 10^3 J/mol\), \(R = 8.314 J/(mol*K)\), \(T_1 = 623 K\) and \(k_2 = 8.80 × 10^{-4} s^{-1}\) to calculate \(T_2\).
04

Convert to Degree Celsius

The solution from Step 3 will yield the temperature in Kelvin. We need to convert this to Degree Celsius by subtracting by 273. This gives the final answer.

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Most popular questions from this chapter

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