Problem 65 - Chemistry Matter and Change

Question

To vaporize 2.00 g of ammonia, 656 calories are required. How many kilojoules are required to vaporize the same mass of ammonia?

Step-by-Step Solution

Answer

The energy required to vaporize 2.00 g of ammonia given as 656 calories can be converted to kilojoules using the following conversion factors: 1 calorie = 4.184 joules and 1 kilojoule = 1000 joules. By performing the conversions, we find that the required energy to vaporize the same mass of ammonia is \(2.745 \, \text{kilojoules}\).

Step 1: Convert calories to joules

We are given the energy required to vaporize ammonia in calories and need to convert it to joules. To do this, we can use the conversion factor: 1 calorie = 4.184 joules Now, multiply the given energy in calories by the conversion factor: Energy in joules = (656 calories) × (4.184 joules/calorie)

Step 2: Calculate the energy in joules

Perform the multiplication from the previous step: Energy in joules = (656 calories) × (4.184 joules/calorie) = 2744.864 joules

Step 3: Convert joules to kilojoules

Now, we need to convert the energy value from joules to kilojoules. To do this, we can use the conversion factor: 1 kilojoule = 1000 joules Now, divide the energy in joules by the conversion factor: Energy in kilojoules = (2744.864 joules) / (1000 joules/kilojoule)

Step 4: Calculate the energy in kilojoules

Perform the division from the previous step: Energy in kilojoules = (2744.864 joules) / (1000 joules/kilojoule) = 2.744864 kilojoules

Step 5: Round the answer

Finally, round the energy value in kilojoules to an appropriate number of significant figures. Since the original energy value in calories had three significant figures (2.00 g), we will round our answer to three significant figures as well: Energy in kilojoules = 2.745 kJ (rounded to three significant figures) Thus, 2.745 kilojoules of energy are required to vaporize the same mass of ammonia as the given 656 calories.

Key Concepts

Calories to Joules

Understanding how to convert calories to joules is essential, especially in scientific and nutrition contexts. A calorie is a unit of energy commonly used to describe fuel or heat energy. It is often used in the context of food energy. To convert calories to joules, you can use the conversion factor:

• 1 calorie = 4.184 joules

In our exercise, we had 656 calories that we needed to convert to joules. We multiply the number of calories by the conversion factor:

• Energy in joules = 656 calories × 4.184 joules per calorie
• This calculation yields 2744.864 joules.

This conversion is crucial when comparing energy values in different units, as the joule is the SI unit of energy.

Joules to Kilojoules

Once energy is in joules, converting it to kilojoules is straightforward and frequently used in chemistry and physics. Joules measure energy in the metric system, and kilojoules, being 1000 joules, provide a more manageable number for larger energy values. To convert joules to kilojoules, use the conversion:

• 1 kilojoule = 1000 joules

To perform this conversion, divide the joules by 1000:

• Energy in kilojoules = 2744.864 joules / 1000
• This calculation results in 2.744864 kilojoules.

Converting to kilojoules makes it easier to communicate and understand energy quantities in scientific discussions.

Significant Figures

Significant figures are crucial in scientific calculations to ensure accuracy and precision. They reflect the precision of a measured or calculated quantity. When using significant figures, you communicate how precise your measurements are. In our example, we started with 656 calories and 2.00 grams — both given to three significant figures.
Applying this precision to our final energy conversion in kilojoules, we need to round our result accordingly. Initially, we have:

• Energy in kilojoules = 2.744864 kJ

Rounding this to three significant figures (matching the precision of the original measurement) gives us:

• Energy in kilojoules = 2.745 kJ

By rounding appropriately, you ensure that your results remain consistent with the accuracy of your input data.

Ammonia Vaporization

Ammonia (NH₃) is a common compound utilized in industrial and chemical processes, and understanding its properties, such as vaporization, is vital. Vaporization refers to the process of a liquid turning into a vapor, an important concept in thermodynamics.
This process requires energy to overcome intermolecular forces and convert the liquid into gas.

• This energy is often measured in calories or joules.
• The given exercise demonstrates that 656 calories are needed to vaporize 2.00 grams of ammonia.

Such conversions and calculations are essential in industrial applications where ammonia is used as a refrigerant or a fertilizer. Efficient energy management during ammonia's vaporization is key to optimizing these processes.