Question

The activation energy Ea for a particular reaction is 50.0 kJ/mol. How
much faster is the reaction at 322 K than at 310.0 K? (R = 8.314 J/mol
•K)

Answer 1

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:

$ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Thus, we plug in Ea, R and the temperatures to obtain:

$ln(\frac{k_2}{k_1} )=-\frac{50,000J/mol}{8.314J/mol*K}(\frac{1}{322K} -\frac{1}{310.0K} ) \\\\ln(\frac{k_2}{k_1} )=0.723$

Now, we use exponential to obtain:

$\frac{k_2}{k_1} =exp(0.723)\\\\\frac{k_2}{k_1}=2.1$

Thus, we infer that the reaction is about two times faster.

Best regards!

Answer 2

We have that at temperature of 325k reaction will be 2.45 times faster than at 310k.

Speed of reaction

Question Parameters:

Generally the Arrhenius  equation  is mathematically given as

$log\frac{k^2}{k^1}=\frac{Ea}{2.303R}(\frac{1/t1}{1/t2})\\\\Therefore\\\\log\frac{k^2}{k^1}=\frac{50e3}{2.3038.314}(\frac{1/310}{1/325})1/k\\\\log\frac{k^2}{k^1}=\frac{2611.35*15}{310*325}\\\\K2=2.45*k1$

Therefore, we infer that

At Temperature of 325k reaction will be 2.45 times faster than at 310k.

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