Question

Which of the following statement is falsc? (1) The number of gram molecules of a substance present in unit volume is known as active mass. (2) Active mass of \(56 \mathrm{~g}\) of \(\mathrm{N}_{2}\) contained in a 2 -litre flask is \(0.1\) mole lit \(1 .\) (3) The attainment of equilibrium state in a gaseous reaction is known by constancy of pressure. (4) There is no effect of pressure on the equilibrium in which all the reactants and products are in liquid state.

Short answer

The false statement is: (2)

Step-by-step solution

Analyze first statement

The first statement defines active mass as the number of gram molecules of a substance present in unit volume. This is correct, as active mass is indeed represented as concentration in mole per liter.

Analyze second statement

We need to calculate the active mass of 56 grams of \(N_2\) in a 2-liter flask. The molar mass of \(N_2\) is 28 g/mol. Therefore, 56 g of \(N_2\) equals \(\frac{56}{28}=2\) moles. The active mass is calculated as: \(\frac{\text{moles}}{\text{volume}}=\frac{2 \text{ moles}}{2 \text{ liters}}=1\) mole/liter. Thus, the second statement, which claims the active mass is 0.1 mole/liter, is false.

Analyze third statement

The third statement claims that the attainment of equilibrium in a gaseous reaction is indicated by constant pressure. This is true as, at equilibrium, the rates of forward and backward reactions are equal, leading to constant pressure in a closed system.

Analyze fourth statement

According to the fourth statement, pressure has no effect on the equilibrium when all reactants and products are in the liquid state. This is true because liquids are nearly incompressible, so changes in pressure do not affect their equilibrium.

Key concepts

Active Mass

Active mass is a key concept in chemical equilibrium. It refers to the number of moles of a substance contained in one liter (or cubic decimeter) of a solution or mixture. In simple words, it's the concentration of a substance in a given volume. The formula to calculate active mass is: \[ \text{Active Mass} = \frac{\text{Number of Moles}}{\text{Volume in Liters}} \] This means if you have a solution of 2 moles of a substance in a 1-liter container, the active mass is 2 moles per liter. Active mass is crucial when studying reactions because it helps determine the rate at which reactants are converted into products. More reactants (higher active mass) generally mean a faster reaction.
To illustrate this with an example taken from the exercise, if you have 56 grams of nitrogen (\text{N}_2) and you want to calculate its active mass in a 2-liter flask:
- First, find the number of moles. The molar mass of \text{N}_2 is 28 g/mol.
- So, 56 grams of \text{N}_2 equals 2 moles (\frac{56}{28} = 2).
- Active mass is then calculated as:
\[ \frac{2 \text{ moles}}{2 \text{ liters}} = 1 \text{ mole per liter} \] This simple calculation shows you how concentration is linked to active mass.

Mole Calculation

Understanding mole calculation is fundamental in chemistry. It helps relate the mass of a substance to the number of particles it contains. This is crucial when dealing with reactions because it lets you quantify reactants and products. Here's a step-by-step on how to calculate moles:
- **Identify the molar mass:** This is the mass of one mole of a given substance, often found in grams per mole (g/mol). For example, the molar mass of \text{N}_2 (Nitrogen) is 28 g/mol.
- **Measure mass:** Find the mass of the substance you have. In the solved exercise, we had 56 grams of \text{N}_2.
- **Use the formula:** Moles are calculated using the formula: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \] Plug in the known values to compute: \[ \frac{56 \text{ grams}}{28 \text{ g/mol}} = 2 \text{ moles} \]
If you wish to find the active mass of these moles in a specific volume, you've already seen how the active mass formula works.
This basic skill is essential because balanced chemical equations work based on moles, helping chemists to compute yields, limiting reagents, and conversions accurately.

Gaseous Reactions

Gaseous reactions are particularly interesting in the study of chemical equilibrium. In gases, particles move freely and rapidly, and their behavior is interconnected with pressure and volume. When studying gaseous reactions, several key factors come into play:
- **Pressure and Volume:** Gases can be compressed or expanded, affecting the reaction rates and equilibrium. The behavior of gases is often explained using the Ideal Gas Law: \[ PV = nRT \] where \text{P} is pressure, \text{V} is volume, \text{n} is moles, \text{R} is the gas constant, and \text{T} is temperature.
- **Equilibrium:** In a gaseous reaction, equilibrium is reached when the rate of the forward reaction equals the rate of the backward reaction. This is often indicated by constancy in pressure for enclosed systems. For example, if you observe the pressure not changing despite the reaction proceeding, it suggests that the system has reached equilibrium.
- **Effect of Pressure:** Pressure changes directly impact reactions involving gases. For instance, increasing pressure by decreasing volume will generally shift the equilibrium towards fewer moles of gas.
In the given exercise, one statement discussed the constancy of pressure as a sign of equilibrium in gaseous reactions. This constant pressure indicates that the system has balanced the rates of the forward and backward reactions.