Question

Rate as a Function of Temperature: The 2 Point Arrhenius Equation

1. When the temperature is changed from 25°C to 46.3°C, the reaction rate doubles. What is the Activation Energy (Ea) in kJ/mol for this reaction?

2. A first order reaction has a half life of 45.0 minutes at 30.0°C. What is the half life (in minutes) when the reaction is heated to 61.0°C? The activation energy for this reaction is 107.0 kJ/mol.

Answer 1

Answer – 1) Wr are given, T₁ = 25^oC + 273 = 298 K , T₂ = 46.3^oC +273 = 319.3 K

Rate, k1 = 1 , so k2 = 2 , Ea = ?

We know the Arrhenius equation

ln k1/k2 = Ea/R * (1/T₂ – 1/ T₁)

so, ln 1/2 = Ea /8.314 J.mol^-1.K^-1 * ( 1/ 319.3 K – 1/ 298)

-0.693 *8.314 J.mol^-1.K^-1 = Ea * -0.00022

So, Ea = -5.76 / -0.00022   

           = 25743.7 J/mol

            = 25.74 kJ/mol

So, the Activation Energy (Ea) in kJ/mol for this reaction is 25.74 kJ/mol

2) We are given, T₁ = 30^oC + 273 = 303 K , T₂ = 61^oC +273 = 334 K

Half-life, t1 = 45 min, half-life, t2 =? , Ea = 107.0 kJ/mol = 107000J/mol

We know the Arrhenius equation

ln t1/2 = Ea/R * (1/T₂ – 1/ T₁)

so, ln 45 min /t2 = 107000 J.mol^-1 /8.314 J.mol^-1.K^-1 * ( 1/ 334 K – 1/ 303 K)

ln 45 min /t2 = -.394

taking anitln from both side

45 min /t2 = 0.0194

So, half-life, t2 = 45 min / 0.0194

                         = 2319 min

The half-life (in minutes) when the reaction is heated to 61.0°C is 2319 min