The hcl equilibrium bond length is 0.127 nm and the v = 0 to v = 1 transition is observed in the infrared at 2886 cm-1.compute the vibrational energy of hcl in its lowest state. compute the classical limit for the stretching of the hcl bond from its equilibrium length in this state. what percent of the equilibrium bond length is this extension

Vibrational Energy Of HCl in the lowest state :

Classical Limit for stretching of HCl bond from its equilibrium length :

Percent of equilibrium Bond Length :

8.58 %

Explanation:

H-Cl bond Length = 0.127 nm =

Frequency from v = 0 to v = 1 is 2886  

Reduced mass

but this has units in amu , to convert it in Kg divide it by

and 1000(to convert gram itno Kg)

Calculation of Force constant :

Here,

k = force constant

Put the value of frequency , reduced mass and calculate for force constant

Solve the left hand side and square it. Then multiply it with reduced mass

k = 475.97 N/m

Calculation of lowest energy

h = planck's constant =

On solving ,

Calculation of Stretching of HCl bond:

Use formula :

here Q = stretching bond length

Put , k = 475.97 N/m , and solve for Q

take square root

Calculation of Percentage extension:

Percentage

Percentage = 8.58 %

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

just did the assignment and the answer was #1 +2 n