The rate constant of a reaction at 32°C is measured to be "0.055 s"^(-1). If the frequency factor is 1.2xx10^13s^-1, what is the activation barrier?

May 27, 2018

${E}_{A} = 84 \textcolor{w h i t e}{l} {\text{kJ"*"mol}}^{- 1}$

Explanation:

The Arrhenius equation states that

$k = A \cdot {e}^{- \frac{\textcolor{p u r p \le}{{E}_{A}}}{R \cdot T}}$

Taking logarithm of both sides gives

$\ln k = \ln A - \frac{\textcolor{p u r p \le}{{E}_{A}}}{R \cdot T}$

Where

• the rate constant of this particular reaction $k = 0.055 \textcolor{w h i t e}{l} {s}^{- 1}$;

• The frequency factor (a temperature-dependent constant ) $A = 1.2 \times {10}^{13} \textcolor{w h i t e}{l} {\text{s}}^{- 1}$ as given in the question;

• The ideal gas constant R=8.314 color(white)(l) color(darkgreen)("J") * color(darkgreen)( "mol"^(-1)) * "K"^(-1);

• Absolute temperature $T = 32 + 273.15 = 305.15 \textcolor{w h i t e}{l} \text{K}$ at which the reaction take place;

• $\textcolor{p u r p \le}{{E}_{A}}$ the activation barrier (a.k.a. activation energy ) the question is asking for

Solve the second equation for $\textcolor{p u r p \le}{{E}_{A}}$:

$\frac{\textcolor{p u r p \le}{{E}_{A}}}{R \cdot T} = \ln A \textcolor{\mathrm{da} r k b l u e}{-} \ln k$

$\textcolor{p u r p \le}{{E}_{A}} = \left(R \cdot T\right) \cdot \left(\ln A \textcolor{\mathrm{da} r k b l u e}{-} \ln k\right)$
$\textcolor{w h i t e}{{E}_{A}} = \left(R \cdot T\right) \cdot \ln \textcolor{\mathrm{da} r k b l u e}{\frac{\textcolor{b l a c k}{A}}{\textcolor{b l a c k}{k}}}$
color(white)(E_A)=8.314 color(white)(l) color(darkgreen)("J") * color(darkgreen)( "mol"^(-1))* color(red)(cancel(color(black)("K"^(-1)))) * 305.15color(white)(l)color(red)(cancel(color(black)("K")))*ln((1.2xx10^13color(red)(cancel(color(black)("s"^(-1)))))/(0.055color(red)(cancel(color(black)("s"^(-1))))))
$\textcolor{w h i t e}{{E}_{A}} = 8.4 \cdot {10}^{4} \textcolor{w h i t e}{l} \textcolor{\mathrm{da} r k g r e e n}{{\text{J") * color(darkgreen)( "mol}}^{- 1}}$

Therefore the activation barrier of this reaction is

$84 \textcolor{w h i t e}{l} \textcolor{b l a c k}{{\text{kJ") * color(darkgreen)( "mol}}^{- 1}}$