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?

1 Answer
May 27, 2018

Answer:

#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))#