Question

The activation energy for the gas phase decomposition of t-butyl bromide is 170. kJ/mol. (CH3)3CBr(CH3)2C=CH2 + HBr The rate constant is 6.16×10-4 s-1 at 538 K. At what temperature is the rate constant 7.39×10-3 s-1?

Answer 1

Answer 2

Solution: Find Temperature for New Rate Constant

Given:

• Activation energy ($E_a$) = 170 kJ/mol = 170,000 J/mol

• Rate constant at $T_1=538\text{K}$: $k_1=6.16\times 10^{-4}{\text{s}}^{-1}$

• New rate constant: $k_2=7.39\times 10^{-3}{\text{s}}^{-1}$

• Gas constant ($R$) = 8.314 J/(mol·K)

Formula (Arrhenius Equation):

$$\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$

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Step-by-Step Calculation:

1. Compute $\ln \left(\frac{k_2}{k_1}\right)$:

   $$\ln \left(\frac{7.39\times 10^{-3}}{6.16\times 10^{-4}}\right)=\ln (12.0)\approx 2.485$$

2. Plug into Arrhenius Equation:

   $$2.485=\frac{170,000}{8.314}(\frac{1}{538}-\frac{1}{T_2})$$$$2.485=20,448(\frac{1}{538}-\frac{1}{T_2})$$

3. Solve for $\frac{1}{T_2}$:

   $$\frac{2.485}{20,448}=\frac{1}{538}-\frac{1}{T_2}$$$$1.215\times 10^{-4}=0.001858-\frac{1}{T_2}$$$$\frac{1}{T_2}=0.001858-1.215\times 10^{-4}=0.001736$$

4. Find $T_2$:

   $$T_2=\frac{1}{0.001736}\approx 576\text{K}$$

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Final Answer:

The rate constant reaches $7.39\times 10^{-3}{\text{s}}^{-1}$ at 576 K.