Question

The running engine of an automobile produces carbon monoxide (CO), a toxic gas, at the rate of about \(188 \mathrm{~g}\) \(\mathrm{CO}\) per hour. A car is left idling in a poorly ventilated garage that is \(6.0 \mathrm{~m}\) long, \(4.0 \mathrm{~m}\) wide, and \(2.2 \mathrm{~m}\) high at \(20^{\circ} \mathrm{C}\). (a) Calculate the rate of CO production in \(\mathrm{mol} / \mathrm{min}\). (b) How long would it take to build up a lethal concentration of \(\mathrm{CO}\) of 1000 ppmv (parts per million by volume)?

Short answer

Rate: 0.112 mol/min; Time to lethal level: 21.1 min.

Step-by-step solution

Rate of CO Production in Grams per Minute

We start by converting the rate of carbon monoxide (CO) production from grams per hour to grams per minute. Given that the engine produces 188 grams of CO per hour, we divide this by 60 (the number of minutes in an hour) to find the rate per minute.\[\text{Rate in } \frac{\text{g}}{\text{min}} = \frac{188 \text{ g/hour}}{60 \text{ min/hour}} = 3.133 \]

Convert Grams per Minute to Moles per Minute

Using the molar mass of CO, which is approximately 28 g/mol, convert the rate from grams per minute to moles per minute.\[\text{Rate in } \frac{\text{mol}}{\text{min}} = \frac{3.133 \text{ g/min}}{28 \text{ g/mol}} = 0.112 \text{ mol/min}\]

Calculate Volume of the Garage

Compute the total volume of the garage using the formula for the volume of a rectangular prism (length × width × height).\[V = 6.0 \text{ m} \times 4.0 \text{ m} \times 2.2 \text{ m} = 52.8 \text{ m}^3\]

Convert Volume to Liters

Since concentrations are often given in parts per million by volume (ppmv), convert the volume from cubic meters to liters (1 m³ = 1000 L).\[V = 52.8 \text{ m}^3 \times 1000 \text{ L/m}^3 = 52800 \text{ L}\]

Calculate Lethal Concentration of CO in Moles

A lethal concentration is given as 1000 ppmv. This indicates that out of every 1,000,000 volumes, 1000 is CO. Calculate the volume of CO in liters, then convert this to moles using the ideal gas law approximation at standard conditions (1 mole = 22.4 L).\[\text{Volume of CO} = \frac{1000}{10^6} \times 52800 \text{ L} = 52.8 \text{ L}\]\[\text{Moles of CO} = \frac{52.8 \text{ L}}{22.4 \text{ L/mol}} = 2.36 \text{ mol}\]

Calculate Time Required to Reach Lethal Concentration

Using the rate of CO production determined in Step 2 and the moles of CO required for a lethal concentration, calculate the time required.\[\text{Time} = \frac{2.36 \text{ mol}}{0.112 \text{ mol/min}} = 21.1 \text{ min}\]

Key concepts

Rate of Production

Understanding the rate of production of carbon monoxide (CO) is crucial when dealing with engines and closed spaces.
The rate is originally given in terms of weight, specifically grams per hour. This must be converted into a timescale more relevant to the problem, such as grams per minute.
In our example, an automobile engine emits CO at a rate of 188 grams per hour.

• To find out how much CO is produced in a minute, divide by 60, converting grams per hour to grams per minute.
• This calculation reveals the production rate as approximately 3.133 grams per minute.

This metric is significant when determining how quickly CO builds up in an environment.

Lethal Concentration

Carbon monoxide is a deadly gas when it accumulates to a lethal concentration.
Lethal concentration is often specified in parts per million by volume (ppmv).
For instance, 1000 ppmv indicates a potentially fatal concentration indoors if sustained.

• Understanding ppmv means acknowledging that 1000 parts of CO exist per million parts of air.
• To comprehend this, consider converting overall garage volume to match this concentration threshold.

Calculating the volume of CO at lethal levels helps us derive how much time exposure becomes dangerous.

Molar Mass of CO

The molar mass of carbon monoxide, CO, is a foundational element in transforming physical quantities.
The molar mass is approximately 28 g/mol.

• This knowledge allows conversions from grams of CO into moles, which are essential for involving gas law principles.
• During calculations, you may be asked to convert CO from grams per minute to moles per minute, which helps facilitate further analysis.

Conversion to moles is critical as it ties the physical rate of production to chemical principles, like stoichiometry and gas laws.

Volume of Garage

The volume of a space like a garage is vital when determining potential CO accumulation.
It allows calculation of how a closed environment interacts with gas production.

• Calculating the volume of a rectangular garage involves multiplying its length, width, and height.
• In this case, the volume amounts to 52.8 cubic meters.
• Subsequently, the volume can be converted to liters, helpful for ppmv usage, resulting in 52,800 liters.

Comprehending the physical space available enables more accurate estimations of gas distribution and containment.

Ideal Gas Law

The ideal gas law, expressed as \(PV = nRT\), is an invaluable tool when dealing with gases like CO.
This fundamental gas equation links the physical properties of gases to their environmental conditions.

• The law aids in determining the number of moles from a given volume under standardized conditions (1 mole = 22.4 Liters).
• Here, it allows conversion from the volume of CO in liters to moles, matching environmental constraints to the gas's properties.
  It relies on consistent temperature and pressure conditions often assumed as standard for simplification.

Grasping the ideal gas law facilitates translations between chemical properties and real-world environments, ensuring safe gas exposure limits.