# If a temperature increase from 19.0 C to 33.0 C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

Jul 10, 2016

The activation energy barrier is 57.1 kJ/mol.

#### 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)|)))" "

${T}_{2} = \left(33.0 + 273.15\right) K = 306.15 K$
${T}_{1} = \left(19.0 + 273.15\right) K = 292.15 K$
${k}_{2} / {k}_{1} = 3$

Now, let's insert the numbers.

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

ln3 = E_"a"/("8.314 J"·color(red)(cancel(color(black)("K")))^"-1""mol"^"-1") (1/(292.15 color(red)(cancel(color(black)("K")))) - 1/(306.15 color(red)(cancel(color(black)("K")))))

$\ln 3 = {E}_{\text{a"/("8.314 J·mol"^"-1") × 1.565 × 10^"-4}}$

E_a = (ln3 × "8.314 J·mol"^"-1")/(1.565 × 10^"-4") = "57 100 J/mol" = "57.1 kJ/mol"