The rate of change in concentration of \(C\) in the reaction; \(2 A+B \longrightarrow 2 C+3 D\), was reported as \(1.0 \mathrm{~mol}\) litre \(^{-1} \mathrm{sec}^{-1} .\) Calculate the reaction rate as well as rate of change of concentration of \(A, B\), and \(D\).

Short Answer

Expert verified
The reaction rate for A is 1.0 mol L^{-1} sec^{-1}, for B is 0.5 mol L^{-1} sec^{-1}, and for D is 1.5 mol L^{-1} sec^{-1}.

Step by step solution

01

Understand the concept of reaction rate

The reaction rate is defined as the change in concentration of a reactant or a product per unit time. For the reaction equation, the stoichiometric coefficients tell us the proportions of reactants to products. To find the rates of change for other substances, we must take into account these ratios.
02

Determine the rate of disappearance of A

According to the stoichiometry of the given reaction, 2 moles of A are used up (disappear) to produce 2 moles of C. Since the rate of change in concentration of C is given as 1.0 mol L^{-1} sec^{-1}, the rate of disappearance of A will be the same because the stoichiometry is 1:1 (considering the 2 moles ratio for both A and C). Thus, the rate for A is 1.0 mol L^{-1} sec^{-1}.
03

Calculate the rate of disappearance of B

For every 2 moles of C produced, only 1 mole of B is required according to the reaction equation. Therefore, the rate of disappearance of B will be half the rate of production of C which is (1.0 / 2) = 0.5 mol L^{-1} sec^{-1}.
04

Determine the rate of formation of D

The reaction produces 3 moles of D for every 2 moles of C produced. This gives us a ratio of 3/2. To find the rate of formation of D, multiply the rate of change of C (1.0 mol L^{-1} sec^{-1}) by 3/2, yielding a rate of 1.5 mol L^{-1} sec^{-1} for D.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Stoichiometry
Chemical stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It relies on the balanced chemical equation, which provides the mole ratio of the reactants and products. For example, in the reaction \(2 A + B \longrightarrow 2 C + 3 D\), the stoichiometric coefficients are 2 for A and C, 1 for B, and 3 for D. These coefficients tell us that 2 moles of A react with 1 mole of B to produce 2 moles of C and 3 moles of D. Understanding this proportion is essential when calculating the reaction rates for each substance involved. \(\)

Stoichiometry allows us to calculate the amount of product formed from a given quantity of reactants and vice versa. It's a foundational concept in chemistry that ensures the conservation of mass and helps in predicting the outcomes of reactions. When dealing with reaction rates, we apply the same stoichiometric principles to relate the rates of disappearance of reactants to the rates of formation of products.
Concentration Change
The concentration change in a chemical reaction refers to how the amount of a substance in a given volume changes over time. It's measured in units such as moles per liter (Molarity, M) per second \((\mathrm{mol\ L^{-1} sec^{-1})}\). In kinetic studies, this change in concentration is used to determine the reaction rate. For instance, if the concentration of product C in our reaction increases by \(1.0 \mathrm{~mol}\) per liter per second, it means that C is being formed at this rate.\(\)

By tracking concentration changes, chemists can understand how fast a reaction is proceeding. A rapid concentration change indicates a fast reaction, while a slow change suggests a slow reaction. Knowing the rates at which concentrations change for each component in a reaction can enable precise control over the process, which is vital in industrial applications and research.
Rate of Disappearance
The rate of disappearance of a reactant in a chemical reaction is the speed at which the reactant is consumed. It is reflected as a decrease in the concentration of the reactant over time. For our example reaction \(2 A + B \longrightarrow 2 C + 3 D\), when we say the rate of disappearance of A is \(1.0 \mathrm{~mol\ L^{-1} sec^{-1}}\), we mean that 1.0 mole of A is being used up every second in every liter of the reaction mixture.\(\)

This rate can be directly related to the rate of formation of products by the mole ratio provided by the chemical equation's stoichiometry. Therefore, the rate of disappearance is critical for understanding not only how quickly a reactant is consumed but also how this consumption drives the formation of the desired products.
Rate of Formation
Conversely, the rate of formation pertains to how quickly a product is made during a chemical reaction. It is quantified by the increase in the concentration of the product over time. Referring back to our reaction, the rate of formation of D was determined to be \(1.5 \mathrm{~mol\ L^{-1} sec^{-1}}\), indicating that for every liter of reaction mixture, 1.5 moles of D are being produced each second.\(\)

The rate of formation and the rate of disappearance are connected through the stoichiometric coefficients of the balanced chemical equation. By understanding these rates, we gauge not only the efficiency of the reaction but also obtain important insights for scaling up reactions for manufacturing, optimizing reaction conditions, and better understanding the reaction mechanism.

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Most popular questions from this chapter

Find the order of reaction for the rate expression rate \(=K[A][B]^{2 / 3}\). Also suggest the units of rate and rate constant for this expression.

For a reaction; \(3 A \longrightarrow\) Products, it is found that the rate of reaction becomes nine times if concentration of \(A\) is increased three times, calculate order of reaction.

A reaction is second order with respect to a reaction. How is the rate of reaction affected if the concentration of the reactant is : (a) doubled, (b) reduced to \(1 / 2 ?\)

\(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposes according to equation; \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) (a) What does \(-\frac{d\left[\mathrm{~N}: \mathrm{C}_{5}\right]}{d t}\) denote? (b) What does \(\frac{d\left[\mathrm{O}_{2}\right]}{d t}\) denote? (c) What is the unit of rate of this reaction?

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) takes place according to I order as $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} $$ Calculate : (a) The rate constant, if instantaneous rate is \(1.4 \times 10^{-6}\) mol litre \(^{-1}\) \(\mathrm{sec}^{-1}\) when concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.04 \mathrm{M}\) (b) The rate of reaction when concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(1.20 \mathrm{M}\) (c) The concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) when the rate of reaction will be \(2.45 \times 10^{-5} \mathrm{~mol}\) litre \(^{-1} \mathrm{sec}^{-\mathrm{i}}\).

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