Answer:
[tex]1.143*10^{3} J/mol[/tex]
Explanation:
Using the Arrhenius equation for the give problem:
[tex]k = A*exp^{\frac{-E_{a} }{RT} }[/tex]
Taking the natural log (ln) of both sides, we have:
[tex]ln k = ln A - \frac{E_{a} }{RT}[/tex]
In the given problem, we have two rate constants at two different temperatures. Thus:
[tex]ln k_{1} = ln A - \frac{E_{a} }{RT_{1} }[/tex] (1)
[tex]ln k_{2} = ln A - \frac{E_{a} }{RT_{2} }[/tex] (2)
Subtracting equation (1) from equation (2), we have:
[tex]ln k_{2} - ln k_{1} = \frac{E_{a} }{RT_{1} } - \frac{E_{a} }{RT_{2} }[/tex] (3)
[tex]k_{1} = 0.0796 L/mol-s[/tex]; [tex]T_{1} = 737°C = 737+273 = 1010 K[/tex]
[tex]k_{2} = 0.0815 L/mol-s[/tex]; [tex]T_{2} = 947°C = 737+273 = 1220 K[/tex]
Therefore, equation (3) becomes:
[tex]ln 0.0815 - ln 0.0796 = E_{a}[\frac{1}{8.314*1010} - \frac{1}{8.314*1220}][/tex]
-2.507 - (2.531) = Ea*[0.00012 - 0.000099]
Ea = 0.024/0.000021 = 1142.86 J/mol
The activation energy of the reaction in scientific notation is [tex]1.143*10^{3} J/mol[/tex]
