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Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the reaction equation 2 H 2 O 2 ( aq ) ⟶ 2 H 2 O ( l ) + O 2 ( g ) 2H2O2(aq)⟶2H2O(l)+O2(g) The activation energy for this reaction is 75 kJ ⋅ mol − 1 . 75 kJ⋅mol−1. In the presence of a metal catalyst, the activation energy is lowered to 49 kJ ⋅ mol − 1 . 49 kJ⋅mol−1. At what temperature would the non‑catalyzed reaction need to be run to have a rate equal to that of the metal‑catalyzed reaction at 25 ∘ C?

Respuesta :

Answer:

The value of temperature for unanalysed reaction is 456.1 K

Explanation:

Step 1: Data given

The activation energy of uncatalyzed reaction is 75.0 kJ/mol.

The activation energy of catalyzed reaction is 49.0 kJ/mol.

The temperature for catalyzed reaction is  25 °C

Step 2:  The balanced equation

2 H2O2(aq) ⟶ 2H2O(l) + O2(g)

Step 3: According to Arrhenius equation:

k = Ae ^ (-Ea/RT)

⇒ with k = the rate constant

⇒ with Ea = the activation energy = 49 kJ/mol

⇒ with R = the universal gas constant = 8.314 J/mol*K

⇒ with T = the temperature  = 298 K

For uncatalyzed reaction, the rate constant is calculated as:

k(uncatalyzed) = Ae ^ (-75000/8.314 J/mol*K * T)

k(uncatalyzed) = Ae ^ (-9020.93/T)

For catalyzed reaction, the rate constant is calculated as:

k(catalyzed) = Ae ^ (-Ea(catalyzed)/RT))

 ⇒ with Ea(catalyzed) = 49000 J/mol

 ⇒ with T = 298 K

k(catalyzed) = Ae ^ (-49000J/(8.314J/mol *K * 298K))

k(catalyzed) = Ae ^(-19.78)

k(catalyzed) = k(uncatalyzed)

Ae^(-19.78) = = Ae ^ (-9020.93/T)

-19.78 = -9020.93/T

T = -9020.93/-19.78

T = 456.1 K

The value of temperature for unanalysed reaction is 456.1 K

tifanihayyu

At temperature 456K (183°C) the non‑catalyzed reaction need to be run to have a rate equal to that of the metal‑catalyzed reaction at 25 ∘ C

Explanation:

Hydrogen peroxide decomposes spontaneously to yield water and oxygen gas according to the reaction equation 2 H 2 O 2 ( aq ) ⟶ 2 H 2 O ( l ) + O 2 ( g ) 2H2O2(aq)⟶2H2O(l)+O2(g) The activation energy for this reaction is 75 kJ ⋅ mol − 1 . 75 kJ⋅mol−1. In the presence of a metal catalyst, the activation energy is lowered to 49 kJ ⋅ mol − 1 . 49 kJ⋅mol−1. At what temperature would the non‑catalyzed reaction need to be run to have a rate equal to that of the metal‑catalyzed reaction at 25 ∘ C?

Non catalytic reaction is the chemical phenomenon completes without involving any influencing material such catalyst. Non catalytic reactions are 2 types: Non Catalytic Homogeneous Reactions and Non Catalytic Heterogeneous Reactions

The balanced equation is:

[tex]2 H2O2[/tex]  ----> [tex]2 H2O + 1 O2[/tex]

Arrhenius equation is a formula for the temperature dependence of reaction rates

[tex]k = Ae^{\frac{-Ea}{RT} }[/tex], where

[tex]k =[/tex] rate constant

[tex]A =[/tex] pre-exponential factor

[tex]E_{a} =[/tex] activation energy (in the same units as R*T)

[tex]R =[/tex] universal gas constant

[tex]T =[/tex] absolute temperature (in Kelvin)

The rate constant for uncatalyzed reaction

[tex]k(uncatalyzed) = Ae^{\frac{ -75000}{8.314 J/mol*K * T}}\\k(uncatalyzed) = Ae^{\frac{-9020.93}{T} }[/tex]

The rate constant for catalyzed reaction

[tex]k(catalyzed) = Ae^{\frac{-Ea(catalyzed)}{RT} }[/tex]

[tex]Ea(catalyzed) = 49000 J/mol[/tex] with [tex]T = 298 K[/tex]

[tex]k(catalyzed) = Ae^{\frac{ -49000 J}{(8.314J/mol *K * 298K)}}\\k(catalyzed) = Ae^{-19.78}[/tex]

Then k(catalyzed) is equal to k(uncatalyzed)

[tex]k(catalyzed) = k(uncatalyzed)\\Ae^{-19.78} = Ae^{\frac{-9020.93}{T} }\\-19.78 = \frac{-9020.93}{T} \\T = \frac{-9020.93}{-19.78} \\T = 456.1 K = 183 C[/tex]

Learn more about Arrhenius equation https://brainly.com/question/13264890

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