Question

Consider a molecule that can be in one of two different conformation states A or B. These states are two different arrangements of the atoms: e.g., in state B, one part of the molecule could be rotated about a bond with respect to the rest of the molecule. Assume the energies of states A and B are 4e-21 and 8e-21 J respectively. At room temperature, T = 298 K, what is the relative likelihood of the molecule being found in state B vs state A? Submit Answer Tries 0/99 If the temperature decreases by 25 K to T = 273 K, what is the relative likelihood of the molecule being found in state B vs state A? Submit Answer Tries 0/99 If the temperature increases by 100 K to T = 398 K, what is the relative likelihood of the molecule being found in state B vs state A? Submit Answer Tries 0/99

Answer 1

Solutions e → According to collect from the data let us anume that the equations are d → Now we have to consider that the denta - To consider the molecule of the two different conforme to be state ArA let us assume the engines of state A&B to be Ae-21* . 8e-21T are respectively the temparataire T= 298k. → The relative likelihood of the molecule being found in Stahovs State A . Do , B alow AE = axiol ? EB = 8X1021 A EA = 4X1021 How we have to conte the equations are one PLE) eup ( +€/+RT) -@ 86 (en +AE) :" cup ("Content) ze semp (-4/417) w int] var) PALEN) - te exp (-48/637) s op en daax zone) 3 - 37.80*.

- Irux 10-21 it is temparature decreased to be ask to T = 2736 € - The relative likelihood of the molecules being found in state B is stake A Po (En+AE) eup (-41) => cop late * x273) PALEA - 34.58% he have to calculate the temperature at 390k EBT - The Temparature * it is the relative found in to state B increase by the look to T=398k likelihood of the molecules of usstated BBC (A +00) 787) >.04829-3 » UR:2737. To lea) = eng ( → increases They are are the atemparature loves