Answer :
We start with a glass tube that is 87.0 cm long. One end is plugged with material moistened by ammonia, and the opposite end by material moistened by hydrochloric acid. When the gases diffuse into the tube, they react according to
[tex]$$
NH_3(g) + HCl(g) \longrightarrow NH_4Cl(s).
$$[/tex]
Because the two gases diffuse at different rates, the point where they first meet (and the white ring of ammonium chloride forms) is not at the center of the tube.
The rate of diffusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law). Let the diffusion rate of ammonia be [tex]$v_{NH_3}$[/tex] and that of HCl be [tex]$v_{HCl}$[/tex]. Then
[tex]$$
\frac{v_{NH_3}}{v_{HCl}} = \sqrt{\frac{M_{HCl}}{M_{NH_3}}},
$$[/tex]
where the molar masses are given as:
- [tex]$M_{NH_3} = 17.0\text{ g/mol}$[/tex],
- [tex]$M_{HCl} = 36.5\text{ g/mol}$[/tex].
Thus,
[tex]$$
\frac{v_{NH_3}}{v_{HCl}} = \sqrt{\frac{36.5}{17.0}} \approx 1.4653.
$$[/tex]
Now, let [tex]$x$[/tex] be the distance (in cm) from the ammonia-moistened plug to the reaction point. Then the HCl must travel the remaining distance, [tex]$87.0 - x$[/tex] cm, from its end of the tube. Since they start at the same time and meet at the same moment, the time traveled by each gas is the same. The distances the gases travel are directly proportional to their speeds:
[tex]$$
\frac{x}{87.0 - x} = \frac{v_{NH_3}}{v_{HCl}}.
$$[/tex]
Substitute the ratio:
[tex]$$
\frac{x}{87.0 - x} = 1.4653.
$$[/tex]
To solve for [tex]$x$[/tex], multiply both sides by [tex]$(87.0 - x)$[/tex]:
[tex]$$
x = 1.4653 (87.0 - x).
$$[/tex]
Distribute the right-hand side:
[tex]$$
x = 1.4653 \times 87.0 - 1.4653 \, x.
$$[/tex]
Combine like terms:
[tex]$$
x + 1.4653 \, x = 1.4653 \times 87.0,
$$[/tex]
[tex]$$
x(1 + 1.4653) = 1.4653 \times 87.0.
$$[/tex]
That gives
[tex]$$
x = \frac{1.4653 \times 87.0}{1 + 1.4653}.
$$[/tex]
Computing the denominator:
[tex]$$
1 + 1.4653 \approx 2.4653,
$$[/tex]
and the numerator:
[tex]$$
1.4653 \times 87.0 \approx 127.5.
$$[/tex]
Thus,
[tex]$$
x \approx \frac{127.5}{2.4653} \approx 51.7 \text{ cm}.
$$[/tex]
Therefore, the reaction occurs approximately 51.7 cm from the ammonia-moistened plug.
[tex]$$
NH_3(g) + HCl(g) \longrightarrow NH_4Cl(s).
$$[/tex]
Because the two gases diffuse at different rates, the point where they first meet (and the white ring of ammonium chloride forms) is not at the center of the tube.
The rate of diffusion of a gas is inversely proportional to the square root of its molar mass (Graham’s law). Let the diffusion rate of ammonia be [tex]$v_{NH_3}$[/tex] and that of HCl be [tex]$v_{HCl}$[/tex]. Then
[tex]$$
\frac{v_{NH_3}}{v_{HCl}} = \sqrt{\frac{M_{HCl}}{M_{NH_3}}},
$$[/tex]
where the molar masses are given as:
- [tex]$M_{NH_3} = 17.0\text{ g/mol}$[/tex],
- [tex]$M_{HCl} = 36.5\text{ g/mol}$[/tex].
Thus,
[tex]$$
\frac{v_{NH_3}}{v_{HCl}} = \sqrt{\frac{36.5}{17.0}} \approx 1.4653.
$$[/tex]
Now, let [tex]$x$[/tex] be the distance (in cm) from the ammonia-moistened plug to the reaction point. Then the HCl must travel the remaining distance, [tex]$87.0 - x$[/tex] cm, from its end of the tube. Since they start at the same time and meet at the same moment, the time traveled by each gas is the same. The distances the gases travel are directly proportional to their speeds:
[tex]$$
\frac{x}{87.0 - x} = \frac{v_{NH_3}}{v_{HCl}}.
$$[/tex]
Substitute the ratio:
[tex]$$
\frac{x}{87.0 - x} = 1.4653.
$$[/tex]
To solve for [tex]$x$[/tex], multiply both sides by [tex]$(87.0 - x)$[/tex]:
[tex]$$
x = 1.4653 (87.0 - x).
$$[/tex]
Distribute the right-hand side:
[tex]$$
x = 1.4653 \times 87.0 - 1.4653 \, x.
$$[/tex]
Combine like terms:
[tex]$$
x + 1.4653 \, x = 1.4653 \times 87.0,
$$[/tex]
[tex]$$
x(1 + 1.4653) = 1.4653 \times 87.0.
$$[/tex]
That gives
[tex]$$
x = \frac{1.4653 \times 87.0}{1 + 1.4653}.
$$[/tex]
Computing the denominator:
[tex]$$
1 + 1.4653 \approx 2.4653,
$$[/tex]
and the numerator:
[tex]$$
1.4653 \times 87.0 \approx 127.5.
$$[/tex]
Thus,
[tex]$$
x \approx \frac{127.5}{2.4653} \approx 51.7 \text{ cm}.
$$[/tex]
Therefore, the reaction occurs approximately 51.7 cm from the ammonia-moistened plug.
