•Calculate the k(rate constant) at different temperature for the following reaction and plot of ln(k) versus 1000/T then compare with the experimental values.
reaction ：
method：
MP2/aug-cc-pVTZ
CH4 +OH․→ H2O + CH3․
398k
%chk=CH4.chk
#MP2/aug-cc-pVTZ opt freq temperature=398
CH4+OH
0 2
C -2.47771751 -0.09616027 0.64416443
H -2.12106309 -1.10497027 0.64416443
H -2.12104467 0.40823792 1.51781593
H -2.12104467 0.40823792 -0.22948708
H -3.54771751 -0.09614709 0.64416443
H -0.57021235 -1.82730581 0.70404986
O -1.57885227 -2.03986339 0.61433758
800k
%chk=CH4.chk
#MP2/aug-cc-pVTZ opt freq temperature=800
CH4+OH
0 2
C -2.47771751 -0.09616027 0.64416443
H -2.12106309 -1.10497027 0.64416443
H -2.12104467 0.40823792 1.51781593
H -2.12104467 0.40823792 -0.22948708
H -3.54771751 -0.09614709 0.64416443
H -0.57021235 -1.82730581 0.70404986
O -1.57885227 -2.03986339 0.61433758
NH3 +OH․→ H2O + NH2․
398k
%chk=nh3.chk
#MP2/aug-cc-pVTZ opt freq temperature=398
nh3+OH
0 2
N -3.74452968 1.48168734 -1.41220800
H -3.41120778 0.53887425 -1.41220800
H -3.41119057 1.95308752 -0.59571126
H -3.41119057 1.95308752 -2.22870473
H -2.16136864 2.58131694 -2.91333407
O -3.12032054 2.54782592 -2.88350790
800k
%chk=nh3.chk
#MP2/aug-cc-pVTZ opt freq temperature=800
nh3+OH
0 2
N -3.74452968 1.48168734 -1.41220800
H -3.41120778 0.53887425 -1.41220800
H -3.41119057 1.95308752 -0.59571126
H -3.41119057 1.95308752 -2.22870473
H -2.16136864 2.58131694 -2.91333407
O -3.12032054 2.54782592 -2.88350790
