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?

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