Question

A certain reaction has activation energy of 50 kJ/mol. The rate constant at 25 degree Celsius is 0.0039 min-1.what is the value of the rate constant at 75 degree Celsius?​

Answer 1

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Answer 2

Given: $E_{a} =50kJ/mol$. $K_{1}=0.0039min^{-1}$ at $25^{o} C$.

To find: rate constant at $75^{o} C$.

Step-by-step method:

Step 1 of 2

The Arrhenius equation is $K=Ae^{\frac{-E_{a} }{RT} }$.

It can be integrated as $lnK=lnA-\frac{E_{a}}{RT}$.

Two different rate constants can be calculated by the formula:

$ln\frac{K_{1} }{K_{2}} =-\frac{E_{a}}{RT_{1} }+\frac{E_{a}}{RT_{2} }$

$ln\frac{K_{1} }{K_{2}} =-\frac{E_{a}}{R }[\frac{1}{T_{2} }-\frac{1}{T_{1} } ]$

Step 2 of 2

Substituting the values we will get:

$ln\frac{0.0039}{K_{2}} =-\frac{50}{8.31 }[\frac{1}{348 }-\frac{1}{298} ]\\log\frac{0.0039}{K_{2}} =-2.303[\frac{50}{8.31 }][\frac{1}{348 }-\frac{1}{298} ]\\log\frac{0.0039}{K_{2}} =0.00668\\\frac{0.0039}{K_{2}}=1.0155\\K_{2}=0.00384min^{-1}$

The rate constant at $75^{o} C$ is $K_{2}=0.00384min^{-1}$.