Question

The gas phase reaction 2 N2O5(g) → 4 NO2(g) + O2(g) has an activation energy of 103 kJ/mol, and the first order rate constant is 1.35×10-5 min-1 at 266 K. What is the rate constant at 296 K?

Answer 1

Answer:  The rate constant at 296 K is $1.51\times 10^{-3}min^{-1}$

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at 266 K = $1.35\times 10^{-5}min^{-1}$

$K_2$ = rate constant at 296 K = ?

$Ea$ = activation energy for the reaction = 103 kJ/mol = 103000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 266 K

$T_2$ = final temperature = 296 K

Now put all the given values in this formula, we get

$\log (\frac{K_2}{1.35\times 10^{-5}})=\frac{103000}{2.303\times 8.314J/mole.K}[\frac{1}{266}-\frac{1}{296}]$

$\log (\frac{K_2}{1.35\times 10^{-5}})=2.049$

$K_2=1.51\times 10^{-3}min^{-1}$

Thus the rate constant at 296 K is $1.51\times 10^{-3}min^{-1}$