Answer:
Time taken is 9.468 min
Solution:
As per the question:
Temperature, [tex]T_{1} = 500^{\circ}C[/tex] = 273 + 500 = 773 K
Temperature, [tex]T_{1} = 450^{\circ}C[/tex] = 273 + 450 = 723 K
Activation energy, [tex]E_{A} = 50\ kcal/mol[/tex]
Time, t = 1 min
Now,
To calculate the time, [tex]t_{2}[/tex] at [tex]T_{1} = 450^{\circ}C[/tex]
We know that:
[tex]1\ cal = 4.184\ kJ[/tex]
Thus
[tex]E_{A} = 50\times 4.184\times 10^{3} kcal/mol = 2.09\times 10^{5}\ J/mol[/tex]
Now, arranging the Arrhenius eqn in terms of time:
[tex]ln\frac{t_{2}}{t_{1}} = -\frac{E_{a}}{R}(\frac{1}{T_{1}} - \frac{1}{T_{2}})[/tex]
[tex]ln\frac{t_{2}}{1} = -\frac{2.09\times 10^{5}}{8.314}(\frac{1}{773} - \frac{1}{723})[/tex]
[tex]lnt_{2} = 2.248[/tex]
Taking anti log
[tex]t_{2} = e^{2.248} = 9.468\ min[/tex]
