# The activation energy of a certain reaction is 49.4 kJ/mol. At 20 degrees C, the rate constant is .0130 s^-1. What is the rate constant at 100 degree C?

Aug 8, 2016

The rate constant is $\text{1.00 s"^"-1}$.

#### Explanation:

The Arrhenius equation gives the relation between temperature and reaction rates:

color(blue)(|bar(ul(color(white)(a/a) k = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "

where

$k$ = the rate constant
$A$ = the pre-exponential factor
${E}_{\text{a}}$ = the activation energy
$R$ = the Universal Gas Constant
$T$ = the temperature

If we take the logarithms of both sides, we get

$\ln k = \ln A - {E}_{\text{a}} / \left(R T\right)$

Finally, if we have the rates at two different temperatures, we can derive the expression

color(blue)(|bar(ul(color(white)(a/a) ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "

${k}_{1} = \text{0.0130 s"^"-1}$
${E}_{\text{a" = "49.4 kJ/mol" = "49 400 J/mol}}$
$R = \text{8.314 J·K"^"-1""mol"^"-1}$
${T}_{2} = \text{100 °C" = "373.15 K}$
${T}_{1} = \text{20 °" = "293.15 K}$

Now, let's insert the numbers.

$\ln \left({k}_{2} / {k}_{1}\right) = {E}_{\text{a}} / R \left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)$

ln(k_2/k_1) = ("49 400" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1")))) (1/(293.15 color(red)(cancel(color(black)("K")))) - 1/(373.15 color(red)(cancel(color(black)("K")))))

ln(k_2/k_1) = 5942× 7.313 × 10^"-4" = 4.346

$\left({k}_{2} / {k}_{1}\right) = {e}^{4.346} = 77.14$

k_2 = 77.14k_1 = 77.14 × "0.0130 s"^"-1" = "1.00 s"^"-1"