Question

In the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+3 \mathrm{D},\) reactant \(\mathrm{A}\) is found to disappear at the rate of \(6.2 \times 10^{-4} \mathrm{M} \mathrm{s}^{-1}.\) (a) What is the rate of reaction at this point? (b) What is the rate of disappearance of \(\mathrm{B}\) ? (c) What is the rate of formation of D?

Short answer

The rate of reaction is \(3.1 \times 10^{-4}\) M/s, the rate of disappearance of B is \(-3.1 \times 10^{-4}\) M/s, and the rate of formation of D is \(9.3 \times 10^{-4}\) M/s.

Step-by-step solution

Determine the Rate of Reaction

The rate of reaction is the rate of disappearance of any one of the reactants or the rate of appearance of any one of the products. Given the rate of disappearance of reactant A (-6.2 x 10^-4 M/s), the rate of the reaction is simply this rate divided by the stoichiometric coefficient of A in the balanced equation, i.e., 2. The equation for determining this is \(-\frac{1}{2}\) \(\times Rate_{a}\) which gives us - (-6.2 x 10^-4) / 2 = 3.1 x 10^-4 M/s.

Determine the Rate of Disappearance of B

The rate of disappearance of B is the rate of reaction multiplied by the stoichiometric coefficient of B. As there's no coefficient for B presented in the balanced equation, it means that coefficient for B is 1. The equation for determining the disappearance rate of B is \(-Rate_{reaction}\).

Determine the Rate of Formation of D

The rate at which D is formed is equal to the rate of reaction multiplied by the stoichiometric coefficient of D in the balanced equation. The stoichiometric coefficient of D is 3, so we can use the equation \(Rate_{reaction}\) \(\times 3\) to solve.

Key concepts

Stoichiometry

Stoichiometry is like the recipe for a chemical reaction. It tells us how many parts of each substance are needed and what you will get in return. In chemistry, these parts are called moles. The coefficients in a balanced equation give us these mole ratios. For example, in the reaction \(2 \mathrm{A} + \mathrm{B} \longrightarrow \mathrm{C} + 3 \mathrm{D}\), the coefficients "2" for \(\mathrm{A}\) and "3" for \(\mathrm{D}\) are crucial. Here’s why they are important:

• They help calculate how much of a reactant is needed or how much product is formed.
• These coefficients are used to relate the rates of various reactions, known as stoichiometric relationships.

Understanding stoichiometry is important because it helps in predicting the outcomes of reactions and ensuring that you have the right proportions of reactants to get the desired product.

Rate of Disappearance

The rate of disappearance measures how fast a reactant is being used up in a reaction. It’s like watching the levels of a liquid in a container go down. In a chemical equation, it's important to remember the stoichiometric coefficients because they determine how fast each reactant disappears relative to each other. In the example given with \(2 \mathrm{A} + \mathrm{B} \longrightarrow \mathrm{C} + 3 \mathrm{D}\):

• Reactant \(\mathrm{A}\) disappears at a rate of \(6.2 \times 10^{-4} \mathrm{M/s}\). Given that \(\mathrm{A}\) has a coefficient of 2, the rate of disappearance must be adjusted by this factor.
• Reactant \(\mathrm{B}\), which has a coefficient of 1 (though typically not written, it implies one unit), would disappear at the same rate as the overall reaction, assuming no competing reactions.

Therefore, knowing these rates helps to understand the dynamics of the entire reaction, how quickly the reactants are used up, and impacts the rate at which the products are formed.

Rate of Formation

The rate of formation is all about how quickly products appear in a reaction. It is as if the products slowly come into existence during the reaction process. Using the balanced chemical equation and stoichiometry, you can figure out the rate at which each product is formed by using the overall reaction rate as a reference. Here’s how it works for \(\mathrm{D}\) in the example reaction \(2 \mathrm{A} + \mathrm{B} \longrightarrow \mathrm{C} + 3 \mathrm{D}\):

• The stoichiometric coefficient for \(\mathrm{D}\) is 3, indicating that for every unit of reaction, three units of \(\mathrm{D}\) are formed.
• This means if the rate of the reaction is found to be \(3.1 \times 10^{-4} \mathrm{M/s}\), the rate of formation of \(\mathrm{D}\) would be three times this rate.

Understanding the rate of formation is crucial for any chemical process because it shows how quickly the end products are generated, which you might need to control in industrial processes or experiments.