Chemistry, 27.07.2019 04:10, winterrs12

The volume of gas at 20°c and 1.00 bar that adsorbed on a cold surface was 1.52 cm3 at 56.4 kpa and 2.77 cm3 at 108 kpa. find the equilibrium constant, and volume of a monolayer.

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answered: ayoismeisalex

Explanation:

According to Langmuir isotherm,

$\theta = \frac{k \times P}{1 + k \times P}$

where, $\theta$ = fraction coverage by gas molecules

k = rate constant

P = partial pressure of gas

Also, $\frac{V}{V_{monolayer}} = \frac{k \times P}{1 + k \times P}$

or $\frac{V_{monolayer}}{V} = \frac{1 + k \times P}{kP}$

$\frac{V_{monolayer}}{V} = \frac{1}{kP} + 1$

k = $\frac{\frac{V}{V_{monolayer}}}{P + (1 - \frac{V}{V_{monolayer}})}$

Now, equation $P_{1}$ and $P_{2}$ for $V_{monolayer}$ as follows.

$\frac{\frac{V_{1}}{V_{monolayer}}}{P + (1 - \frac{V_{1}}{V_{monolayer}})}$ = $\frac{\frac{V_{2}}{V_{monolayer}}}{P + (1 - \frac{V_{2}}{V_{monolayer}})}$

$\frac{P_{1}(V_{monolayer} - V_{1}}{V_{1}}$ = $\frac{P_{2}(V_{monolayer} - V_{2}}{V_{2}}$

$V_{monolayer} = \frac{P_{1} - P_{2}}{\frac{P_{1}}{V_{1}}} - \frac{P_{2}}{V_{2}}$ .......... (1)

As it is given that $P_{1}$ is 56.4 kPa, $P_{2}$ is 108 kPa, $V_{1}$ is $1.52 cm^{3}$ and $V_{2}$ is $2.77 cm^{3}$ .

Therefore, putting these values into equation (1) as follows.

$V_{monolayer} = \frac{P_{1} - P_{2}}{\frac{P_{1}}{V_{1}}} - \frac{P_{2}}{V_{2}}$

= $\frac{56.4 kPa - 108 kPa}{\frac{56.4 kPa}{1.52 cm^{3}}} - \frac{108 kPa}{2.77 cm^{3}}$

= 27.44 $cm^{3}$

Thus, we can conclude that volume of monolayer is 27.44 $cm^{3}$ .

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