The rate constant of a reaction at 32°C is 0.060/s. If the frequency factor is 3.1 × 10 15 s –1 , what is the activation barrier?

The rate constant of a reaction at 32 degrees Celsius is 0.060 seconds inverse. If the frequency factor is 3.1 times 10 to the 15 seconds inverse, what is the activation barrier? Alright. So in this question, they're giving us temperature, they're giving us our rate constant, We have our frequency factor, and activation barrier is just activation energy. Now there's 2 formulas that could help us isolate e a. The first one is just simply the Arrhenius equation, which is k equals a times e to the negative e a over r t. Now to isolate that e a, we'd have to do some manipulation of this formula, which has in it already inverse of the natural log, which is e. Now the manipulations we have to do gives us a second equation, so let's just not worry about the Arrhenius equation and just use the second equation. The second equation is our linear form of the Arrhenius equation. So that's l n k equals negative e a over r times 1 over t plus l n of a. Here we have l n of 0.060 equals negative e a over r which is 8.314, and then one over, so temperature we add 273.15, which is 305.15 Kelvin, plus l n of 3.1 times 10 to the 15th. So when we do l n of 0.060, we get negative 2.81341 and this equals negative e a over, so here 8.314 and 305.15 can multiply together. When they do, they give us 25 37.0171, and this is still plus l n of 3.1 times 10 to 15. Here we're going to subtract ln3.1 times 10 to 15 from both sides. So when we do that, we're going to get negative 38.4836 equals negative e a over 2537.0171. Multiply both sides by 2537.0171, So what we're gonna get here initially is negative 97 633.5 equals negative e a. Here divide both sides by negative one, so we're gonna get e a equals, and here we're just gonna use 2 sig figs, so 9.8 times 10 to the 4 joules per mole. Here, activation energy is in joules because r is 8.314 joules over moles times k. Okay. So its units dictate what units for e a or activation energy would be. So this would be our final answer for the activation barrier.

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15. Chemical Kinetics

Arrhenius Equation

General Chemistry

15. Chemical Kinetics

Arrhenius Equation

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Multiple Choice

The rate constant of a reaction at 32°C is 0.060/s. If the frequency factor is 3.1 × 10¹⁵ s^–1, what is the activation barrier?

9.1 x 10⁴J/mol

1.0 x 10⁴J/mol

9.8 x 10⁴J/mol

8.3 x 10⁴J/mol

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Verified step by step guidance

Start by understanding the Arrhenius equation, which is used to relate the rate constant (k) of a reaction to the temperature (T) and the activation energy (Ea). The equation is: k = A * e^(-Ea/(RT)), where A is the frequency factor, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.

Convert the given temperature from Celsius to Kelvin. Since the temperature is given as 32°C, add 273.15 to convert it to Kelvin: T = 32 + 273.15 = 305.15 K.

Rearrange the Arrhenius equation to solve for the activation energy (Ea). Take the natural logarithm of both sides to get: ln(k) = ln(A) - Ea/(RT).

Substitute the known values into the rearranged equation. You have k = 0.060/s, A = 3.1 × 10^15 s^–1, R = 8.314 J/mol·K, and T = 305.15 K. Plug these values into the equation: ln(0.060) = ln(3.1 × 10^15) - Ea/(8.314 * 305.15).

Solve for Ea by isolating it on one side of the equation. This involves calculating the difference between the natural logarithms and multiplying by RT to find the activation energy in J/mol.