Question

Chemists commonly use a rule of thumb that an increase of 10K in temperature doubles the rate of a reaction What must the activation energy be for this statement to be true for a temperature increase from 59 to 69 C? kmol Activation energy

Answer 1

Answer:

Step 1: Explanation:

The Arrhenius equation allows us to calculate activation energies if the rate constant is known.as activation energy term E_a increases, the rate constant k decreases and therefore the rate of reaction decreases

K =Ae^-E_a^/RT

or, ln ( k₂/k₁) = -Ea/R(1/T₂-1/T₁)

where, k represents the rate constant, E_a is the activation energy, R is the gas constant (8.3145 J/K mol), and T is the temperature expressed in Kelvin. A is known as the frequency factor, having units of L mol^-1 s^-1

Step 2: Calculation of Activation Energy

Temperature (T₁) =(59+273.15)K = 332.15 K

Temperature (T₂) = ( 69+273.15)K = 342.15 K

It is assumed that the rate constant at 332.15 K is K₁ = K and as the temperature increased to 342.15 K , the rate constant becomes double i.e K₂ = 2K

on substituting the values in above equation

ln ( k₂/k₁) = -E_a/R(1/T₂-1/T₁)

ln ( 2k / k ) =((-E_a) / 8.314 J/ mol^-1 K^-1 )) × (1/342.15 K -1/332.15 K)

0.6931471806 =((-E_a) / 8.314 J/ mol^-1 K^-1 )) × (-8.799321766 ×.10^-5 K^-1 )

[ here two negative hence both will cancel out each other ]

E_a = ( 0.6931471806  × 8.314 J/ mol^-1 K^-1 ) /   (8.799321766 ×.10^-5 K^-1 )

E_a = 65491.7 J/mol

Hence the activation energy in kJ will be

E_a = 65491.7 J/mol  × ( 1 kJ / 1000 J ) = 65.5 kJ / mol [ note: 1 kJ = 100J ]