Instantaneous vs average rates; molar gas volumes - Chemistry - HKDSE

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Chemistry Study Notes: Reaction Rates & Molar Gas Volumes

Hey everyone! Welcome to your study notes for a really important part of Chemistry: Rate of Reaction. In this chapter, we're going to zoom in on two key ideas. First, we'll figure out how to measure how fast a reaction is going at any given moment, just like a speedometer in a car. Then, we'll learn a super useful shortcut for dealing with gases in chemical reactions. Understanding these concepts is crucial for mastering stoichiometry and kinetics questions in your exams. Let's dive in!

Part 1: How Fast? Average vs. Instantaneous Rates

Imagine you're on a road trip from Hong Kong Island to the New Territories. Your overall journey might take an hour. But at different points, you might be stuck in traffic or cruising on the highway. Chemistry is the same! Reactions don't always happen at a steady speed. We need two ways to describe their speed, or 'rate'.

What is the Rate of Reaction?

In simple terms, the rate of reaction is how quickly reactants are used up, or how quickly products are formed. We usually measure this by looking at the change in concentration, volume, or mass over a period of time.

The formula we often use is:

$$ \text{Rate} = \frac{\text{Change in amount (concentration, volume, mass)}}{\text{Change in time}} $$

We can see this visually on a graph that plots the amount of a substance against time. Typically, the curve is steepest at the beginning (when the reaction is fastest) and becomes flatter as the reactants are used up.

The 'Road Trip' Speed: Average Rate of Reaction

The average rate tells you the overall speed of the reaction over a specific time interval. It's like saying your average speed for the whole road trip was 60 km/h, even though you were sometimes faster or slower.

How to find it from a graph:

Pick two points on the curve, (t₁, y₁) and (t₂, y₂).
Calculate the change in the y-axis (amount) and the x-axis (time).
Divide the change in amount by the change in time. This is the gradient of the line connecting the two points.

Formula:

$$ \text{Average Rate between } t_1 \text{ and } t_2 = \frac{y_2 - y_1}{t_2 - t_1} $$

Example: Let's say we are measuring the volume of CO₂ produced. At 10 seconds, we have 20 cm³ of gas. At 40 seconds, we have 50 cm³ of gas.

The average rate between 10s and 40s is:

$$ \text{Average Rate} = \frac{50 \text{ cm}^3 - 20 \text{ cm}^3}{40 \text{ s} - 10 \text{ s}} = \frac{30 \text{ cm}^3}{30 \text{ s}} = 1.0 \text{ cm}^3\text{s}^{-1} $$

The 'Speedometer' Reading: Instantaneous Rate of Reaction

The instantaneous rate is the exact speed of the reaction at one specific moment in time. It's like looking at your car's speedometer to see you're going at exactly 80 km/h right now.

How to find it from a graph:

This is a bit trickier, but don't worry! You just need to draw a tangent.

Find the exact time point you're interested in on the x-axis (e.g., t = 20 seconds).
Go up to the curve.
Carefully draw a straight line that just touches the curve at that single point. This line is called a tangent. Make it as long as you can for better accuracy!
Calculate the gradient (slope) of this tangent line. This gradient is the instantaneous rate.

Formula (for the tangent line):

$$ \text{Instantaneous Rate at time t} = \text{Gradient of the tangent at t} = \frac{\Delta y}{\Delta x} $$

A very important instantaneous rate is the initial rate. This is the rate right at the start of the reaction (at time t = 0). It's the fastest the reaction will be because the concentration of reactants is at its highest. To find it, you simply draw the tangent at t = 0.

A Quick Review Box

Average Rate: The rate over a TIME INTERVAL. Calculated using two points ON the curve.
Instantaneous Rate: The rate at a specific MOMENT in time. Calculated from the gradient of the TANGENT to the curve at that point.

Key Takeaway for Reaction Rates

The difference between average and instantaneous rate is like the difference between your average speed on a whole journey versus your speed at one exact moment. For your exams, you must be able to calculate both from a graph. Remember: instantaneous rate = tangent gradient!

Part 2: Molar Volume of Gases - The Chemist's Shortcut

Working with gases can be a pain. Measuring their mass is hard, but measuring their volume is easy! Luckily, chemists discovered a fantastic relationship that makes calculations with gases much simpler.

What is Molar Volume?

Let's quickly refresh our memory. A mole is just a specific number of particles ($$6.02 \times 10^{23}$$). The cool thing about gases is that at the same temperature and pressure, one mole of ANY gas takes up the same amount of space (volume).

The molar volume of a gas is the volume occupied by one mole of that gas under a specific set of conditions.

For HKDSE, the most important condition is Room Temperature and Pressure (r.t.p.).

Temperature: 25°C (or 298 K)
Pressure: 1 atmosphere (atm)

The Magic Number: 24 dm³

At r.t.p., one mole of any gas—whether it's hydrogen (H₂), oxygen (O₂), or carbon dioxide (CO₂)—has a volume of 24 dm³. (Remember, 1 dm³ is the same as 1 litre or 1000 cm³).

This is the key concept:
1 mole of any gas at r.t.p. = 24 dm³

Analogy: Think of it like a standard-sized box. It doesn't matter if you fill the box with feathers or bricks; the box's volume is always the same. Similarly, at r.t.p., the 'box' for one mole of any gas always has a volume of 24 dm³.

Calculations with Molar Volume

We can relate moles, volume, and molar volume with a simple formula. A great way to remember it is with a formula triangle.

The Formula Triangle:
Imagine a triangle with "Volume of Gas (in dm³)" at the top, and "Number of Moles" and "Molar Volume (24 dm³)" at the bottom.

From this, we get two key equations:

$$ \text{Volume of gas (dm}^3\text{)} = \text{Number of moles} \times 24 $$ $$ \text{Number of moles} = \frac{\text{Volume of gas (dm}^3\text{)}}{24} $$

Let's try a problem! Step-by-step.

Question: What is the volume of hydrogen gas produced at r.t.p. when 1.2 g of magnesium reacts completely with excess hydrochloric acid? (Molar mass of Mg = 24.3 g mol⁻¹)

Step 1: Write the balanced chemical equation.

$$ \text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} $$

Step 2: Find the moles of the substance you know.
We know the mass of magnesium.

$$ \text{Moles of Mg} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.2 \text{ g}}{24.3 \text{ g mol}^{-1}} = 0.0494 \text{ mol} $$

Step 3: Use the mole ratio from the equation to find the moles of the gas.
From the equation, the ratio of Mg to H₂ is 1:1.

So, Moles of H₂ = Moles of Mg = 0.0494 mol.

Step 4: Convert the moles of gas to volume using the molar volume.
Now we use our magic number!

$$ \text{Volume of H}_2 = \text{Moles of H}_2 \times 24 \text{ dm}^3\text{mol}^{-1} $$ $$ \text{Volume of H}_2 = 0.0494 \text{ mol} \times 24 \text{ dm}^3\text{mol}^{-1} = 1.19 \text{ dm}^3 $$

So, 1.19 dm³ of hydrogen gas is produced. See? It's that simple!

Common Mistakes to Avoid

Forgetting to use a balanced equation to find the mole ratio. This is the most common error!
Using the molar volume (24 dm³) for liquids, solids, or aqueous solutions. It only works for gases.
Using 24 dm³ when the conditions are NOT r.t.p. The question must state that the reaction is at r.t.p.
Mixing up units. If your volume is given in cm³, remember to divide by 1000 to convert it to dm³ before using the formula.

Key Takeaway for Molar Volume

The molar volume of a gas is your best friend for stoichiometry problems involving gases. Just remember the magic number 24 dm³ per mole at r.t.p. and follow the steps: find moles, use the ratio, then find the volume. You've got this!