Question

The half-life for the reaction below was determined to be 2.14 × 10^4 s at 800 K. The value of the half-life is independent of the inital concentration of N2O present. The activation energy of the reaction is 270.00 kJ/mol.
N2O------>N2+O


A)What would be the half-life at 1150.66 K?

____________s

Answer 1

Answer:

0.0907 s

Explanation:

This an Arrhenius equation problem, so you relate the half-life with the kinetic constant of the reaction in order to calcule the same thermodynamic parameters at another temperature.

To calcule the kinetic constant of the reaction you need to know the order of it, look closely to the sentence "The value of the half-life is independent of the inital concentration of N2O present." the only order independent from the initial concentration of reagents is first order, so you can calculate K at 800 K, using:

$k(800 K)=\frac{ln(2)}{t_{1/2}}= \frac{ln(2)}{2.14 * 10^{4} s}}=3.239*10^{-5}s^{-1}}$

Now you can use Arrhenius equation to calcule K at 1150.66 K

$ln(\frac{k1}{k2} )=-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} )$

$k2= k1*exp(-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} ))=3.239*10^{-5}s^{-1}*exp(-\frac{270 000 J/mol }{8.314 J/mol *k }(\frac{1}{1150.66K} - \frac{1}{800K} ))=7.639 s^{-1}$

Then calculate the new half-life:

$t_{1/2} =\frac{ln(2)}{7.639s^{-1}}=0.0907 s$