rate = k [ A ] m [ B ] n {\displaystyle {\mbox{rate}}=k[A]^{m}[B]^{n}\,\!}
order of reaction = m + n {\displaystyle {\mbox{order of reaction}}=m+n\,\!}
k = A e − E a R T {\displaystyle k=Ae^{-{\frac {E_{a}}{RT}}}}
ln ( k ) = ln ( A e − E a R T ) {\displaystyle \ln {(k)}=\ln {\left(Ae^{-{\frac {E_{a}}{RT}}}\right)}}
ln ( k ) = ln ( A ) + ln ( e − E a R T ) {\displaystyle \ln {(k)}=\ln {(A)}+\ln {\left(e^{-{\frac {E_{a}}{RT}}}\right)}}
ln ( k ) = ln ( A ) + − E a R T {\displaystyle \ln {(k)}=\ln {(A)}+-{\frac {E_{a}}{RT}}}
ln ( k ) = ln ( A ) − E a R T {\displaystyle \ln {(k)}=\ln {(A)}-{\frac {E_{a}}{RT}}}
ln ( k ) = − E a R T + ln ( A ) {\displaystyle \ln {(k)}=-{\frac {E_{a}}{RT}}+\ln {(A)}}
ln ( k ) = − E a R 1 T + ln ( A ) {\displaystyle \ln {(k)}=-{\frac {E_{a}}{R}}{\frac {1}{T}}+\ln {(A)}}
straight-line graphs are always of the form y = m x + c {\displaystyle {\mbox{straight-line graphs are always of the form }}y=mx+c\,\!}
therefore, the dependent variable y is ln ( k ) {\displaystyle {\mbox{therefore, the dependent variable }}y{\mbox{ is }}\ln {(k)}\,\!}
the gradient m is − E a R {\displaystyle {\mbox{the gradient }}m{\mbox{ is }}-{\frac {E_{a}}{R}}\,\!}
the independent variable x is 1 T {\displaystyle {\mbox{the independent variable }}x{\mbox{ is }}{\frac {1}{T}}\,\!}
the y-intercept c is ln ( A ) {\displaystyle {\mbox{the y-intercept }}c{\mbox{ is }}\ln {(A)}\,\!}
![{\displaystyle {\mbox{rate}}=k[A]^{m}[B]^{n}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c8f3cdd83e421d7ff9724bd3ed86a3d54d48be9)
