Question

Challenge If the average reaction rate for the reaction, expressed in moles of \(\mathrm{HCl}\) formed, is \(0.0050 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s},\) what concentration of \(\mathrm{HCl}\) would be present after 4.00 \(\mathrm{s}\) ?

Short answer

The concentration of HCl after 4 seconds can be found using the formula: final concentration = initial concentration + (reaction rate × time), with the initial concentration being 0. Given the average reaction rate of 0.0050 mol/L·s, the final HCl concentration after 4 seconds is \(0 + (0.0050 \ \text{mol/L} \cdot \text{s} \times 4.00 \ \text{s}) = 0.0200 \ \text{mol/L}\).

Step-by-step solution

Identify the given information and find the formula

We are given the average reaction rate: 0.0050 mol/L·s and the time: 4.00 s. We will use the formula: final concentration = initial concentration + (reaction rate × time) to find the concentration of HCl after 4 seconds. Since there is no HCl formed initially, we can consider the initial concentration to be 0.

Apply the formula

We will now substitute the given values into the formula: Final concentration of HCl = 0 + (0.0050 mol/L·s × 4.00 s)

Calculate the final concentration of HCl

Performing the calculations, we get: Final concentration of HCl = 0 + (0.0050 mol/L·s × 4.00 s) = 0.0200 mol/L The concentration of HCl formed after 4 seconds is 0.0200 mol/L.

Key concepts

Reaction Rate Calculation

Understanding reaction rate calculation is crucial for studying chemical reactions. The reaction rate tells you how fast a reactant is being consumed or how quickly a product is forming. In chemical terms, it's often expressed as the change in concentration of a reactant or product per unit time, usually moles per liter per second ((mol/L•s)).

For instance, the given problem states an average reaction rate of 0.0050 mol/L•s for the formation of HCl. This implies that each second, there are 0.0050 moles of HCl being formed in each liter of solution. When calculating the total amount of product formed over a period, we multiply this rate by the time elapsed. Since there's often no initial product in a fresh reaction mixture, if we're given that the reaction rate is constant, the calculation simplifies to just multiplying the reaction rate by the time. This straightforward approach helps students avoid confusion, focusing on the core concept of how reaction rates translate into actual quantities of substances in a given period.

Chemical Concentration

Chemical concentration is a measure of how much of a given substance is present in a mixture. In chemistry, concentration is commonly expressed as moles per liter ((mol/L)), which tells us the amount of a solute in a given volume of solution. The concentration can change over time during a reaction as reactants turn into products.

When confronted with problems regarding concentration, always pay attention to whether you're being asked for the initial, intermediate, or final concentration. This will direct you to either use, solve for, or calculate the respective value as needed. Also, understanding stoichiometry, the mathematical relationship between reactants and products in a reaction, is integral when adjustments in reaction ratios come into play. For students, incorporating step-by-step strategies, like identifying given information first, simplifies tackling concentration-related problems.

Moles per Liter

The term 'moles per liter' is a unit of concentration specifically used in the field of chemistry. A mole is a standard unit that measures the amount of substance, and it's defined as containing exactly 6.02214076×10²³ elementary entities (like atoms or molecules), a number known as Avogadro's number. Thus, when we speak of moles per liter, we are quantifying the number of molecules or atoms of a solute in a liter of solution.

This unit often confuses students, who might be misled by the word 'liter,' which is a measure of volume, not the number of entities. However, by remembering that a mole is simply a count of particles, similar to a dozen being 12 of something, it quickly becomes more intuitive. Reinforcing this concept by practical examples and exercises, such as the given problem regarding HCl formation, helps solidify the students' understanding and application of this fundamental unit in chemistry.