An oxygen molecule is adsorbed on a patch of surface (see sketch at right). This patch is known to contain adsorption sites. The molecule has enough energy to move from site to site, so it could be on any one of them. Suppose part of the surface becomes inaccessible, so that only adsorption sites are now available for the molecule. Calculate the change in entropy. Round your answer to significant digits, and be sure it has the correct unit symbol.

This question is incomplete, the complete question is;

An oxygen  (O₂) molecule is adsorbed on a patch of surface (see sketch at right). This patch is known to contain 324 adsorption sites.

The  (O₂) molecule has enough energy to move from site to site, so it could be on any one of them.

Suppose part of the surface becomes inaccessible, so that only 400 adsorption sites are now available for the molecule.

Calculate the change in entropy.

Round your answer to 3 significant digits, and be sure it has the correct unit symbol.

the change in entropy is 2.91  x 10⁻²³ J/K

Explanation:

Given the data in the question;

For an ideal gas,

Entropy of a molecule, S = K ln(W)

where K  is Boltzmann constant (1.38065 x 10⁻²³ J/K)

W is Equivalent micro states of the molecule

( W = 324   and W = 400 )

S - S = K ln(W) - K ln(W )

ΔS = K ln(W / W )

we substitute

ΔS = (1.38065 x 10⁻²³ J/K) × ln( 400 / 324 )

ΔS = 2.9093  x 10⁻²³ J/K

ΔS = 2.91  x 10⁻²³ J/K

Therefore,  the change in entropy is 2.91  x 10⁻²³ J/K

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).

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