Question

For the reaction \(3 A(g) \stackrel{k}{\longrightarrow} B(g)+C(g), k\) is \(10^{-14} \mathrm{~L} / \mathrm{mol} . \mathrm{min}\). if \([A]=0.5 M\) then the value of \(-\frac{d[A]}{d t}\) (in \(M \mathrm{~s}^{-1}\) ) is: (a) \(7.5 \times 10^{-5}\) (b) \(3 \times 10^{-4}\) (c) \(2.5 \times 10^{-5}\) (d) none of these

Short answer

\(-\frac{d[A]}{d t} = 1.25 \times 10^{-5} \mathrm{M/s} \), which is none of these.

Step-by-step solution

Write down the rate law

According to the given chemical reaction, the rate of the reaction can be represented using the rate law, which, for a reaction of the type \(nA \rightarrow products\), is given by the formula \(rate = k[A]^n\). Since the stoichiometric coefficient of A in the given reaction is 3, the rate can be written as \(rate = k[A]^3\), where \(k\) is the rate constant and \([A]\) is the concentration of reactant A.

Determine the rate of disappearance of A

The rate of disappearance of A is given by the negative of the rate of the reaction because A is being consumed. Thus, \(-\frac{d[A]}{dt} = k[A]^3\). Here, \(k = 10^{-14} \mathrm{L/mol \cdot min}\) and \([A] = 0.5 \mathrm{M}\).

Calculate the rate of disappearance of A

Substitute the value of \(k\) and \([A]\) into the rate law equation to find \(-\frac{d[A]}{dt}\): \(-\frac{d[A]}{dt} = 10^{-14} \times (0.5)^3\). Calculate the result to find \(-\frac{d[A]}{dt}\) in \(\mathrm{L/mol \cdot min}\).

Convert the units from mol/L⋅min to M/s

To convert the rate from \(\mathrm{L/mol \cdot min}\) to \(M/s\) (M is mol/L), multiply the result from Step 3 by \((1 \mathrm{min})/(60 \mathrm{s})\), since there are 60 seconds in a minute: \(-\frac{d[A]}{dt} \times \frac{1}{60}\).

Find the correct answer

Calculate the final value for \(-\frac{d[A]}{dt}\) after unit conversion and compare it with the given options to identify the correct answer.

Key concepts

Chemical Reaction Rates

Understanding chemical reaction rates is key to mastering kinetics, a fundamental concept in chemistry. The rate of a chemical reaction refers to how fast the reactants are converted into products. This speed can be affected by various factors including concentration, temperature, and the presence of a catalyst. In the context of this exercise, we look at the rate at which reactant A disappears in a gas-phase reaction.

When chemists talk about the rate of reaction, they often refer to the rate of change in concentration of a reactant or product per unit time. For a balanced chemical equation, this can be represented as the negative change in the concentration of a reactant over time, hence the term rate of disappearance. In our exercise, this is denoted by \( -\frac{d[A]}{dt} \), signaling that A is being consumed as the reaction progresses. The negative sign indicates a decrease in concentration over time.

Rate Constant Calculation

The rate constant, symbolized by k, is a proportionality factor in the rate law of a chemical reaction and provides the relationship between the reactant concentration and the rate of reaction. Its value is determined experimentally and remains constant at a given temperature.

In the given problem, we are provided with the rate constant k for the reaction 3 A(g) \(\stackrel{k}{\longrightarrow}\) B(g) + C(g). The provided value of \(k = 10^{-14} \text{ L/mol} \cdot \text{min}\) indicates how rapidly the reaction occurs under the current conditions. To apply the rate constant to our rate law expression \(rate = k[A]^3\), we insert the known values for k and the concentration of A to calculate the rate of the reaction.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It's the backbone of predicting how much of each substance is needed or produced. The stoichiometric coefficients in the balanced equation give the ratio in which reactants are consumed and products formed.

In our rate law problem, the stoichiometry of the reaction 3 A(g) \(\stackrel{k}{\longrightarrow}\) B(g) + C(g) highlights that three moles of A are required to form the products B and C. When interpreting the rate law, the stoichiometric coefficient of reactant A, which is 3, tells us that the rate is third order with respect to A. This means that the rate depends on the concentration of A raised to the third power. Thus, small changes in the concentration of A can greatly affect the reaction rate.

Units Conversion in Chemistry

Units conversion in chemistry is a fundamental skill to ensure measurements are accurate and consistent across different systems of units. In kinetics, rates are usually expressed in terms of concentration change per unit time, with common units such as \(M/s\) (molarity per second) or \(L/mol \cdot min\).

Consistency in units is crucial when calculating concentration and reaction rates. As seen in this problem, we start with a rate constant given in \(L/mol \cdot min\), and to find the rate of disappearance of A in molarity per second (\(M/s\)), we need to convert from minutes to seconds. This involves simple arithmetic, dividing the value by 60, as there are 60 seconds in a minute. This step is pivotal as it allows us to express the rate in the standard unit of molarity per second, which is more commonly used in scientific calculations and helps in comparing the rates of different reactions.