# A certain reaction with an activation energy of 165 kJ/mol was run at 455 K and again at 475 K. What was the ratio of f at the higher temperature to f at the lower temperature?

Jul 12, 2016

$\left({f}_{2} / {f}_{1}\right) = 6.27$

#### Explanation:

I presume that you are using $f$ as a symbol for the rate constant.

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

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

where

$f$ = 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

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

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

color(blue)(|bar(ul(color(white)(a/a) ln(f_2/f_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "

${E}_{\text{a" = "165 kJ/mol" = "165 000 J/mol}}$
${T}_{2} = \text{475 K}$
${T}_{1} = \text{455 K}$

Now, let's insert the numbers.

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

ln(f_2/f_1) = ("165 000" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1")))) (1/(455 color(red)(cancel(color(black)("K")))) - 1/(475 color(red)(cancel(color(black)("K")))))

$\ln \left({f}_{2} / {f}_{1}\right) = \text{19 846" × 9.254 × 10^"-5} = 1.837$

$\left({f}_{2} / {f}_{1}\right) = {e}^{1.837} = 6.27$