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Science Chemistry CHEM:ATOM FOC 2E CL (TEXT)

Activation energy for the given reaction. Rate constant of the reaction at 300K. Solution: E a = 19.5 K J m o l R a t e c o n s t a n t k a t 300 K = 1.89 × 10 7 M - 1 s - 1 a t T Explanation: Concept: We are asked to find the reaction’s activation energy and rate constant at 300K, which is calculated using the Arrhenius equation. According to the Arrhenius equation, the slope of a plot l n k versus 1 T is equal to - E a R , where E a is the activation energy and R is the universal gas constant with unit J m o l . K . We need to convert the T and k values from the given table to 1 T and ln ⁡ k respectively. We predict a positive value for E a because activation energy is the energy needed to overcome the barrier. Also, the value is in the magnitude of 10 7 , if we consider trend for the values of k at increasing temperatures in the given data. Formula: E a = - s l o p e × R ln ⁡ k 1 k 2 = E a R 1 T 2 - 1 T 1 Given: We are given values of the rate constant k as a function of absolute temperature. N O 2 ( g ) + O 3 ( g ) → N O 3 ( g ) + O 2 ( g ) T ( K ) k ( M - 1 s - 1 ) 203 4.14 × 10 5 213 7.30 × 10 5 223 1.22 × 10 5 233 1.96 × 10 6 243 3.02 × 10 6 Calculations: Expanding the data table to include columns of ln ⁡ k and 1 T yields the above table. A plot of ln ⁡ k versus 1 T gives us the slope, which is - 2450.6 K . We use the values of the slope and R to calculate the value of E a . T ( K ) 1 T ( K - 1 ) k ( M - 1 s - 1 ) ln ⁡ k 203 0.004926 4.14 × 10 5 12.93362125 213 0.004695 7.30 × 10 5 13.50079981 223 0.004484 1.22 × 10 5 11.71177632 233 0.004292 1.96 × 10 6 14.48845503 243 0.004115 3.02 × 10 6 14.92076739 E a = - s l o p e × R E a = - - 2348.9 K × 8.314 J m o l . K = 1.95 × 10 4 J m o l = 19.5 k J m o l Calculation of rate constant of the reaction at 300 K : This can be calculated using modified Arrhenius equation: ln ⁡ k 1 k 2 = E a R 1 T 2 - 1 T 1 For this, consider T 1 = 300 K ; k 1 = ? ; T 2 = 243 K ; k 2 = 3.02 × 1 0 6 M - 1 s - 1 Plugging values for k 2 , T 1 , a n d T 2 in the modified Arrhenius equation and solving the equation for k 1 : ln ⁡ k 1 k 2 = E a R 1 T 2 - 1 T 1 l n k 1 3.02 × 1 0 6 M - 1 s - 1 = 1.95 × 10 4 J m o l 8.314 J m o l . K 1 243 K - 1 300 K l n k 1 3.02 × 1 0 6 M - 1 s - 1 = 2348.9 K - 1 × 7.8189 × 10 - 4 K l n k 1 3.02 × 1 0 6 M - 1 s - 1 = 1.836588477366 k 1 3.02 × 1 0 6 M - 1 s - 1 = 6.27509407769 k 1 = 6.2750940776 × 3.02 × 1 0 6 k 1 = 1.89 × 10 7 M - 1 s - 1 Thus, the rate constant for the above reaction at 300 K is 1.89 × 10 7 M - 1 s - 1 . Conclusion: The activation energy is the height of a barrier that has to be overcome before a reaction can proceed. We can calculate the activation energy for a given reaction if values of rate constants at specified temperatures are given.

Activation energy for the given reaction. Rate constant of the reaction at 300K. Solution: E a = 19.5 K J m o l R a t e c o n s t a n t k a t 300 K = 1.89 × 10 7 M - 1 s - 1 a t T Explanation: Concept: We are asked to find the reaction’s activation energy and rate constant at 300K, which is calculated using the Arrhenius equation. According to the Arrhenius equation, the slope of a plot l n k versus 1 T is equal to - E a R , where E a is the activation energy and R is the universal gas constant with unit J m o l . K . We need to convert the T and k values from the given table to 1 T and ln ⁡ k respectively. We predict a positive value for E a because activation energy is the energy needed to overcome the barrier. Also, the value is in the magnitude of 10 7 , if we consider trend for the values of k at increasing temperatures in the given data. Formula: E a = - s l o p e × R ln ⁡ k 1 k 2 = E a R 1 T 2 - 1 T 1 Given: We are given values of the rate constant k as a function of absolute temperature. N O 2 ( g ) + O 3 ( g ) → N O 3 ( g ) + O 2 ( g ) T ( K ) k ( M - 1 s - 1 ) 203 4.14 × 10 5 213 7.30 × 10 5 223 1.22 × 10 5 233 1.96 × 10 6 243 3.02 × 10 6 Calculations: Expanding the data table to include columns of ln ⁡ k and 1 T yields the above table. A plot of ln ⁡ k versus 1 T gives us the slope, which is - 2450.6 K . We use the values of the slope and R to calculate the value of E a . T ( K ) 1 T ( K - 1 ) k ( M - 1 s - 1 ) ln ⁡ k 203 0.004926 4.14 × 10 5 12.93362125 213 0.004695 7.30 × 10 5 13.50079981 223 0.004484 1.22 × 10 5 11.71177632 233 0.004292 1.96 × 10 6 14.48845503 243 0.004115 3.02 × 10 6 14.92076739 E a = - s l o p e × R E a = - - 2348.9 K × 8.314 J m o l . K = 1.95 × 10 4 J m o l = 19.5 k J m o l Calculation of rate constant of the reaction at 300 K : This can be calculated using modified Arrhenius equation: ln ⁡ k 1 k 2 = E a R 1 T 2 - 1 T 1 For this, consider T 1 = 300 K ; k 1 = ? ; T 2 = 243 K ; k 2 = 3.02 × 1 0 6 M - 1 s - 1 Plugging values for k 2 , T 1 , a n d T 2 in the modified Arrhenius equation and solving the equation for k 1 : ln ⁡ k 1 k 2 = E a R 1 T 2 - 1 T 1 l n k 1 3.02 × 1 0 6 M - 1 s - 1 = 1.95 × 10 4 J m o l 8.314 J m o l . K 1 243 K - 1 300 K l n k 1 3.02 × 1 0 6 M - 1 s - 1 = 2348.9 K - 1 × 7.8189 × 10 - 4 K l n k 1 3.02 × 1 0 6 M - 1 s - 1 = 1.836588477366 k 1 3.02 × 1 0 6 M - 1 s - 1 = 6.27509407769 k 1 = 6.2750940776 × 3.02 × 1 0 6 k 1 = 1.89 × 10 7 M - 1 s - 1 Thus, the rate constant for the above reaction at 300 K is 1.89 × 10 7 M - 1 s - 1 . Conclusion: The activation energy is the height of a barrier that has to be overcome before a reaction can proceed. We can calculate the activation energy for a given reaction if values of rate constants at specified temperatures are given.

CHEM:ATOM FOC 2E CL (TEXT)

CHEM:ATOM FOC 2E CL (TEXT)

2nd Edition

ISBN: 9780393284218

Author: Stacey Lowery Bretz, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss

Publisher: WW Norton & Co

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CHEM:ATOM FOC 2E CL (TEXT)

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