Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)

The correct answer is 5.61 × 10⁻⁵.

Explanation:

Based on the given information, let us consider the experimentally determined value of K as 0.000717, with the value of T₁ as 45 degree C, the value of T₁ in Kelvin would be,

T₁ = 273 +45 = 318 K

The value of T₂ is 25 degree C, or (25+273) = 298 K

Based on Arrhenius equation:

ln(K₂/K₁) = -Ea/R(1/T₂ - 1/T₁)

Now putting the values in the equation we get,

ln(K₂/0.000717) = -12070 (1/298 - 1/318)

ln(K₂/0.000717) = -2.54738

K₂/0.000717 = e^-2.54738 = 0.078286

K₂ = 0.078286 × 0.000717

K₂ = 5.61 × 10⁻⁵

c. 4.54 x 10–12 j happens to be correct