When a chemical reaction is carried out in an open container, it means the system is open to the surroundings, and the pressure remains constant. This constant pressure is typically the atmospheric pressure, often denoted as \( p_{ex} \). If the reaction forms a gas, the volume of the system will increase as the gas expands into the surroundings.

Understanding Work Done by a System

Work done in thermodynamics is related to changes in volume against an external pressure. For a system undergoing a volume change against a constant external pressure \( p_{ex} \), the work done ($W$) is given by the formula:

\( W = - p_{ex} \Delta V \)

Where:

• \( p_{ex} \) is the constant external pressure.
• \( \Delta V \) is the change in volume of the system, calculated as \( V_{final} - V_{initial} \).

Why the Negative Sign?

The sign convention for work is important:

• Work done by the system on the surroundings is considered negative (the system expends energy). When a gas expands against external pressure, it is doing work on the surroundings.
• Work done on the system by the surroundings is considered positive (the system gains energy). When the surroundings compress the system, work is done on the system.

In this question, the formation of gas causes the system's volume to increase, meaning the system expands and does work on the surroundings. Therefore, the work done by the system must be negative. Since the volume increases, \( \Delta V \) is positive, so the formula \( W = - p_{ex} \Delta V \) correctly yields a negative value for the work done by the system.

Analyzing the Options

Let's look at the given options in the context of a chemical reaction forming gas in an open container:

• 0: This would only be true if there were no change in volume (\( \Delta V = 0 \)) or if there were no external pressure. Since a gas is formed in an open container, the volume is likely to change, and there is always atmospheric pressure. So, this option is generally incorrect.
• \( p_{ex} \Delta V \): This formula would represent work done *on* the system when its volume decreases, or the magnitude of the work done *by* the system, but it lacks the necessary negative sign to represent work done *by* the system as per convention when volume increases.
• \( - p_{ex} \Delta V \): This formula correctly represents the work done *by* the system when its volume changes by \( \Delta V \) against a constant external pressure \( p_{ex} \). Since gas formation leads to expansion (\( \Delta V > 0 \)), this formula gives a negative value, indicating work done by the system.
• \( V \Delta p_{ex} \): This formula is related to work done in different scenarios (like work done *on* a system during compression at variable volume, or work done on a rigid container if pressure changes). In an open container, the external pressure \( p_{ex} \) is essentially constant, so \( \Delta p_{ex} = 0 \). This option does not apply here.

Based on the definition of pressure-volume work done by a system at constant external pressure, the work done when gas is formed in an open container causing volume expansion is \( - p_{ex} \Delta V \).