Question

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?

Answer 1

`E_A=84color(white)(l)"kJ"*"mol"^(-1)`

Explanation:

The Arrhenius equation states that

`k=A*e^(-(color(purple)(E_A))/(R*T))`

Taking logarithm of both sides gives

`lnk=lnA-(color(purple)(E_A))/(R*T)`

Where

• the rate constant of this particular reaction `k=0.055color(white)(l)s^(-1)`;

• The frequency factor (a temperature-dependent constant ) `A=1.2xx10^13color(white)(l)"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=305.15color(white)(l)"K"` at which the reaction take place;

• `color(purple)(E_A)` the activation barrier (a.k.a. activation energy ) the question is asking for

Solve the second equation for `color(purple)(E_A)`:

`color(purple)(E_A)/(R*T)=lnAcolor(darkblue)(-)lnk`

`color(purple)(E_A)=(R*T)*(lnAcolor(darkblue)(-)lnk)`
`color(white)(E_A)=(R*T)*lncolor(darkblue)(color(black)(A)/color(black)(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))))))`
`color(white)(E_A)=8.4*10^4color(white)(l)color(darkgreen)("J") * color(darkgreen)( "mol"^(-1))`

Therefore the activation barrier of this reaction is

`84color(white)(l)color(black)("kJ") * color(darkgreen)( "mol"^(-1))`