Given that the initial rate constant is 0.0110s−1 at an initial temperature of 30 ∘C , what would the rate constant be at a temperature of 200. ∘C for the same reaction described in Part A? Ea = 31.8 kJ
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 30 ∘C ,...
Given that the initial rate constant is 0.0180s−1 at an initial temperature of 30 ∘C , what would the rate constant be at a temperature of 180. ∘C for the same reaction described in Part A? Express your answer with the appropriate units. (I solved this and got 2.01, which is wrong, but I'm not exactly sure why). (Reaction A was: The activation energy of a certain reaction is 44.3 kJ/mol . At 30 ∘C , the rate constant is...
Given that the initial rate constant is 0.0160s-1 at
an initial temperature of 24 deg C, what would the rate constant be
at a temperature of 100 deg C for the same reaction described in
Part A?
From Part A:
The activation energy of a certain reaction is 31.5 kJ/mol. At
24∘C, the rate constant is 0.0160s−1. T2 = 41C
Part B at an initial temperature of 24 °C, what would the rate constant be at a temperature of 100....
The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is lnk2k1=EaR(1T1−1T2) which is mathmatically equivalent to lnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate constants for a single reaction at two different absolute...
The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is lnk2k1=EaR(1T1−1T2) which is mathmatically equivalent to lnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate constants for a single reaction at two different absolute...
The activation energy of a certain reaction is 33.6 kJ/mol . At 23 ∘C , the rate constant is 0.0150s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Given that the initial rate constant is 0.0150s−1 at an initial temperature of 23 ∘C , what would the rate constant be at a temperature of 120. ∘C for the same reaction described in Part A?
The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is lnk2k1=EaR(1T1−1T2) which is mathmatically equivalent to lnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate constants for a single reaction at two different absolute...
Part A The activation energy of a certain reaction is 34.9 kJ/mol . At 23 ∘C , the rate constant is 0.0110s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Part B Given that the initial rate constant is 0.0110s−1 at an initial temperature of 23 ∘C , what would the rate constant be at a temperature of 120. ∘C for the same reaction described in Part A?
Part A: The activation energy of a certain reaction is 44.9 kJ/mol. At 25 ∘C, the rate constant is 0.0120s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Part B: Given that the initial rate constant is 0.0120s−1 at an initial temperature of 25 ∘C, what would the rate constant be at a temperature of 120. ∘C for the same reaction described in Part A?
A: The activation energy of a certain reaction is 36.8 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 100. ∘C for the same reaction described in Part A?
The activation energy of a certain reaction is 47.9 kJ/molkJ/mol . At 29 ∘C ∘C , the rate constant is 0.0180s−10.0180s−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units. T2= Given that the initial rate constant is 0.0180s−10.0180s−1 at an initial temperature of 29 ∘C ∘C , what would the rate constant be at a temperature of 170. ∘C ∘C for the same reaction described in Part A? k2=