Question

The Brunauer-Emmett-Teller (BET) surface area measurement data for adsorption of \(\mathrm{N_2}\) gas at 77 K on 1.0 g of an adsorbent fits into a straight line, when \(\dfrac{z}{(1-z)V}\) is plotted against \(z\). The slope and intercept of the straight line are \(6 \times 10^{-4}\) mm\(^{-3}\) and \(4 \times 10^{-6}\) mm\(^{-3}\), respectively. The surface area (in m\(^2\) g\(^{-1}\)) of the adsorbent is (rounded off to one decimal place).
(Given: 1 mm\(^3\) of \(\mathrm{N_2}\) gas corresponds to \(2.7 \times 10^{16}\) molecules and each molecule occupies 0.16 nm\(^2\). \(V\) is the volume of gas adsorbed in mm\(^3\), \(z\) is \(p/p_o\), where \(p\) is the pressure of the \(\mathrm{N_2}\) gas and \(p_o\) is equilibrium vapour pressure of liquid \(\mathrm{N_2}\))

Hint:

Slope + intercept of the BET plot equals \(1/V_m\); convert \(V_m\) (mm\(^3\)) to molecules using the given molecules-per-mm\(^3\) figure, then multiply by the area per molecule.

Answer 1

Correct Answer: 7.2

Step 1: Write the BET equation in linear form.
The BET adsorption isotherm, in the linear form used for this plot, is
\[ \frac{z}{(1-z)V} = \frac{1}{V_mC} + \frac{C-1}{V_mC}\,z \] So intercept \(= \dfrac{1}{V_mC}\) and slope \(=\dfrac{C-1}{V_mC}\), where \(V_m\) is the monolayer volume of gas (per gram of adsorbent here, since the sample is 1.0 g) and \(C\) is the BET constant.

Step 2: Add slope and intercept to eliminate \(C\).
\[ \mathrm{slope}+\mathrm{intercept} = \frac{C-1}{V_mC}+\frac{1}{V_mC} = \frac{C}{V_mC} = \frac{1}{V_m} \] \[ V_m = \frac{1}{\mathrm{slope}+\mathrm{intercept}} = \frac{1}{(6\times10^{-4})+(4\times10^{-6})} = \frac{1}{6.04\times10^{-4}} \] \[ V_m = 1655.63\ \mathrm{mm^3} \] This is the monolayer volume of \(\mathrm{N_2}\) gas (at the measurement conditions) adsorbed on the 1.0 g sample.

Step 3: Convert \(V_m\) into the number of molecules in the monolayer.
Given: 1 mm\(^3\) of \(\mathrm{N_2}\) gas corresponds to \(2.7\times10^{16}\) molecules.
\[ N = V_m \times 2.7\times10^{16} = 1655.63 \times 2.7\times10^{16} = 4.4702\times10^{19}\ \mathrm{molecules} \]
Step 4: Convert molecule count to surface area.
Each \(\mathrm{N_2}\) molecule covers \(0.16\ \mathrm{nm^2} = 0.16\times10^{-18}\ \mathrm{m^2}\).
\[ A = N \times 0.16\times10^{-18}\ \mathrm{m^2} = (4.4702\times10^{19})(1.6\times10^{-19}) = 7.1523\ \mathrm{m^2} \]
Step 5: State per-gram surface area.
Since this area belongs to the 1.0 g adsorbent sample used in the measurement, the specific surface area is directly \(7.1523\ \mathrm{m^2\,g^{-1}}\).

Final Answer:
Rounded to one decimal place, the BET surface area is \[ \boxed{7.2\ \mathrm{m^2\,g^{-1}}} \]