Question

The rotational constant for H35cl to observed to be 10.5909 c/m . what are the value of B for H37cl and for 2d35cl

Answer 1

Answer:

the rotational constant for h35cl is observed at 10.5909 cm^-1. what are the values of b for h37cl and 2d35cl - 33803181.

Answer 2

The rotational constant is 5.4425 cm-1.

We are given that rotational constant of H35Cl to be observed to be 10.5909/cm.

Diatomic molecules with rotational energy levels

In this equation, J represents the quantum number for total rotational angular momentum, and B represents the rotational constant, which is related to the molecule's moment of inertia, I = ur2 (u is the reduced mass, and r is the bond length).

Rotational constant B = $\frac{h}{8\pi ^{2}Ic }$

and I = $ur^{2}$

Now, considering the internuclear distance 'r' will be the same,

⇒ r(HCl) = r(DCl)

so, we can derive that, $\frac{B(DCl)}{B(HCl)} = \frac{u * (HCl)}{u*(DCl)}$

⇒ $B(DCl) = \frac{u*(HCl)}{u*(DCl)}*B(HCl)$

Now reduced mass $u = \frac{m_{1}m_{2} }{m_{1}+m_{2} }$ and m = $\frac{M}{N_{A} }$

$u(HCl) = \frac{M(H)}{N_{A} } * \frac{M(Cl)}{N_{A}}/\frac{M(H)}{N_{A}}*\frac{M(Cl)}{N_{A} }$

$u(DCl) = \frac{M(D)}{N_{A} } * \frac{M(Cl)}{N_{A}}/\frac{M(D)}{N_{A}}*\frac{M(Cl)}{N_{A} }$

≅ $\frac{u(HCl)}{u(DCl)} = \frac{M(D)+M(Cl)}{M(H)+M(Cl)} * \frac{M(H)}{M(D)}$

⇒ $\frac{(2+35)g mol^{-1} }{(1+35)g mol^{-1} } * \frac{1g mol^{-1} }{2gmol^{-1} }$

= 37/72

∴ $B(DCl) = \frac{u*(HCl)}{u*(DCl)}*B(HCl)$ = (37/72) *10.5909 $cm^{-1}$ = 5.4425 $cm^{-1}$.

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