Question

In: Chemistry

A) The activation energy of a certain reaction is 48.0 kJ/mol . At 27 ∘C ,...

A) The activation energy of a certain reaction is 48.0 kJ/mol . At 27 ∘C , the rate constant is 0.0120s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

B) Given that the initial rate constant is 0.0120s−1 at an initial temperature of 27 ∘C , what would the rate constant be at a temperature of 170. ∘C for the same reaction described in Part A?

Expert Solution

A) We need to use the Arrhenius equation; the integrated form is given as

ln k₂/k₁ = E_a/R*(1/T₁ – 1/T₂)

where k₁ is the rate of the reaction at absolute temperature T₁ and k₂ is the rate of the reaction at absolute temperature T₂.

Given E_a = 48.0 kJ/mol = (48.0 kJ/mol)*(1000 J/1 kJ) = 48000 J/mol, T₁ = 27°C ≡ (27 + 273) K = 300 K, k₁ = 0.0120 s^-1 and k₂ = 2*0.0120 s^-1 = 0.0240 s^-1, plug in values and obtain

ln (0.0240 s^-1)/(0.0120 s^-1) = (48000 J/mol)/(8.314 J/mol.K)*(1/300 – 1/T₂) K^-1

====> ln (2) = (5773.394275)*(0.003333 – 1/T₂)

====> 0.69315 = (5773.394275)*(0.003333 – 1/T₂)

====> 0.003333 – 1/T₂ = 0.69315/5773.394275 = 0.000120

====> 1/T₂ = 0.003213

====> T₂ = 1/0.003213 = 311.2356 ≈ 311.23

The desired temperature is 311.23 K ≡ (311.23 – 273)°C = 38.23°C (ans).

B) We shall use the Arrhenius equation again.

Given E_a = 48.0 kJ/mol = (48.0 kJ/mol)*(1000 J/1 kJ) = 48000 J/mol, T₁ = 27°C ≡ (27 + 273) K = 300 K, T₂ = 170°C ≡ (170 + 273) K = 443 K and k₁ = 0.0120 s^-1, plug in values and obtain

ln k₂/k₁ = E_a/R*(1/T₁ – 1/T₂)

====> ln k₂/(0.0120 s^-1) = (48000 J/mol)/(8.314 J/mol.K)*(1/300 – 1.443) K^-1

====> ln k₂/(0.0120 s^-1) = (5773.394275)*(0.001076) = 6.2122

====> k₂/(0.0120 s^-1) = exp^(6.2122) = 498.7974

====> k₂ = 498.7974*(0.0120 s^-1) = 5.98556 s^-1 ≈ 5.9856 s^-1 (ans).

queen_honey_blossom answered 2 years ago

a) The activation energy of a certain reaction is 47.0 kJ/mol . At 24 ∘C ,...

a) The activation energy of a certain reaction is 47.0 kJ/mol . At 24 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast? b)Given that the initial rate constant is 0.0190s−1 at an initial temperature of 24 ∘C , what would the rate constant be at a temperature of 100. ∘C for the same reaction described in Part A?

a,The activation energy of a certain reaction is 37.1 kJ/mol . At 30 ∘C , the...

a,The activation energy of a certain reaction is 37.1 kJ/mol . At 30 ∘C , the rate constant is 0.0170s−1. At what temperature in degrees Celsius would this reaction go twice as fast? b,Given that the initial rate constant is 0.0170s−1 at an initial temperature of 30 ∘C , what would the rate constant be at a temperature of 170. ∘C for the same reaction described in Part A?

A. The activation energy of a certain reaction is 42.0 kJ/mol . At 29 ∘C ,...

A. The activation energy of a certain reaction is 42.0 kJ/mol . At 29 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast? B. Given that the initial rate constant is 0.0190s−1 at an initial temperature of 29 ∘C , what would the rate constant be at a temperature of 120. ∘C for the same reaction described in Part A? Please give the correct units too. Thank you!

The activation energy of a certain reaction is 35 kJ/mol . At 20 ∘C, the rate...

The activation energy of a certain reaction is 35 kJ/mol . At 20 ∘C, the rate constant is 0.0130 s−1. At what temperature would this reaction go twice as fast? Express your answer numerically in degrees Celsius

Part A: The activation energy of a certain reaction is 34.4 kJ/mol . At 24 ∘C...

Part A: The activation energy of a certain reaction is 34.4 kJ/mol . At 24 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. Part B: Given that the initial rate constant is 0.0190s−1 at an initial temperature of 24 ∘C , what would the rate constant be at a temperature of 130. ∘C for the same reaction described in Part A? Express your...

Part A The activation energy of a certain reaction is 40.6 kJ/mol . At 20 ∘C...

Part A The activation energy of a certain reaction is 40.6 kJ/mol . At 20 ∘C , the rate constant is 0.0150s−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. Part B Given that the initial rate constant is 0.0150s−1 at an initial temperature of 20 ∘C , what would the rate constant be at a temperature of 110. ∘C for the same reaction described in Part...

A certain reaction has an activation energy of 39.79 kJ/mol. At what Kelvin temperature will the...

A certain reaction has an activation energy of 39.79 kJ/mol. At what Kelvin temperature will the reaction proceed 6.50 times faster than it did at 339 K?

Part A The activation energy of a certain reaction is 39.4kJ/mol . At 30 ?C ,...

Part A The activation energy of a certain reaction is 39.4kJ/mol . At 30 ?C , the rate constant is 0.0190s?1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. Part B Given that the initial rate constant is 0.0190s?1 at an initial temperature of 30 ?C , what would the rate constant be at a temperature of 150 ?C for the same reaction described in Part A? Express...

The activation energy for a reaction is changed from 184 kJ/mol to 59.3 kJ/mol at 600....

The activation energy for a reaction is changed from 184 kJ/mol to 59.3 kJ/mol at 600. K by the introduction of a catalyst. If the uncatalyzed reaction takes about 2653 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor A is constant and assume the initial concentrations are the same

A certain first-order reaction has an activation energy of 53 kJ mol-1. It is run four...

A certain first-order reaction has an activation energy of 53 kJ mol-1. It is run four times, once at 298 K, again at 308 K, a third time at 398 K, and finally at 408 K. Assume that k1 = rate constant at 298 K, k2 = rate constant at 308 K, k3 = rate constant 398 K, and k4 = the rate constant at 408 K. Assuming that all other factors are identical, identify the correct ratios between rate...