Question

The activation energy for the gas phase decomposition of Nitroperoxynitrate is 30.5 kJ/mol. 2N2O7 → 4NO2 + 3O2 The rate constant at 285 K is. 0.127 sec-1. What is the Rate Constant at 270 K? What does the Activation Energy tell us about the compound?

Answer 1

For the given reaction  2N₂O₇ → 4NO₂ + 3O₂

In order to solve this problem, to calculate the rate contant at a particular temperature from the Activation Energy (Ea) values we have the following integrated expression from Arrhenius Equation at different temperature given below :

ln (K₂/K₁) = E_a / R [ (1/T₁) - (1/T₂)]

Given :

Activation Energy (Ea) = 30.5 kJ/mol

Rate Constant , K₁ = 0.127 sec^-1 and T₁ = 285 K

Rate constant , K₂ = ? and T₂ = 270 K

Therefore,

ln (K₂ / 0.127 sec^-1) = 30.5 kJ/mol / 8.314 Jmol^-1K^-1 [(1/285 K) - (1/270 K)]

ln (K₂ / 0.127 sec^-1) = 30500 J/mol / 8.314 Jmol^-1K^-1 [(270-285) / 285x270 K]

ln (K₂ / 0.127 sec^-1) = 30500 J/mol / 8.314 Jmol^-1K^-1 [(-15) / 76950K]

ln (K₂ / 0.127 sec^-1) = - 0.715

K₂ / 0.127 sec^-1 = e^{- 0.715}

K₂ = 0.127 sec^-1 * e^{- 0.715​​​​​​​}

K₂ = 0.489 * 0.127 sec^-1

K₂ = 0.062 sec^-1

• Hence the Rate constant (K₂) at 270 K= 0.062 sec^-1
• The positive Activation Energy for this reaction tells that as the tempreature increases the rate of the reaction also increases which is evident in the above problem i.e. the molecules gains kinetic energy and move faster and faster towards products.