Answer:
About two times faster.
Explanation:
Hello!
In this case, since we need a k2/k1 ratio to figure out the increase in the reaction rate for a chemical process, we need to use the following version of the Arrhenius equation:
[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Thus, we plug in Ea, R and the temperatures to obtain:
[tex]ln(\frac{k_2}{k_1} )=-\frac{50,000J/mol}{8.314J/mol*K}(\frac{1}{322K} -\frac{1}{310.0K} ) \\\\ln(\frac{k_2}{k_1} )=0.723[/tex]
Now, we use exponential to obtain:
[tex]\frac{k_2}{k_1} =exp(0.723)\\\\\frac{k_2}{k_1}=2.1[/tex]
Thus, we infer that the reaction is about two times faster.
Best regards!
We have that at temperature of 325k reaction will be 2.45 times faster than at 310k.
Question Parameters:
Generally the Arrhenius equation is mathematically given as
[tex]log\frac{k^2}{k^1}=\frac{Ea}{2.303R}(\frac{1/t1}{1/t2})\\\\Therefore\\\\log\frac{k^2}{k^1}=\frac{50e3}{2.3038.314}(\frac{1/310}{1/325})1/k\\\\log\frac{k^2}{k^1}=\frac{2611.35*15}{310*325}\\\\K2=2.45*k1\\\\[/tex]
Therefore, we infer that
At Temperature of 325k reaction will be 2.45 times faster than at 310k.
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