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For the reaction \(\mathrm{A} \longrightarrow\) Products, it is found that the rate of reaction increases by a factor of \(6.25\), when the concentration of \(\mathrm{A}\) is increased by a factor of \(2.5 .\) The order of reaction with respect to \(\mathrm{A}\) is (a) \(0.5\) (b) 1 (c) 2 (d) 3

Short Answer

Expert verified
The order of reaction with respect to \( \mathrm{A} \) is 2.

Step by step solution

01

Write the Rate Law Expression

The rate law for the reaction \( \mathrm{A} \rightarrow \) Products is expressed as: \[ \text{Rate} = k [A]^n \] where \( k \) is the rate constant, \( [A] \) is the concentration of \( \mathrm{A} \), and \( n \) is the order of reaction with respect to \( \mathrm{A} \).
02

Setup the Rate Increase Equation

We know that the rate increases by a factor of 6.25 when the concentration of \( \mathrm{A} \) is increased by a factor of 2.5. Substitute these values into the rate equation to get: \[ 6.25 = (2.5)^n \]
03

Solve for the Order \( n \)

Take the logarithm on both sides to solve for \( n \):\[ \log(6.25) = n \cdot \log(2.5) \]Calculate:\[ n = \frac{\log(6.25)}{\log(2.5)} \]Evaluating this, we find:\[ n = 2 \]
04

Confirm the Order of Reaction

The calculated \( n = 2 \) indicates that the order of the reaction with respect to \( \mathrm{A} \) is 2, which matches one of the given options.

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Key Concepts

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

Rate Law
The rate law of a chemical reaction describes how the rate of a reaction depends on the concentration of reactants. It is an expression that provides important insights into the dynamics of the reaction.
For the reaction \( \mathrm{A} \rightarrow \) Products, the rate law can be written as: \[ \text{Rate} = k [A]^n \]Here:
  • \(k\) is the rate constant, a unique value for each reaction at a given temperature.
  • \([A]\) represents the molar concentration of reactant \(A\).
  • \(n\) is the order of the reaction with respect to \(A\), indicating how changes in \([A]\) affect the reaction rate.
The rate law is determined experimentally, as it is specific to each reaction and cannot be deduced from the stoichiometry alone.
Order of Reaction
The order of a reaction is a key part of the rate law equation, represented by the exponent of the concentration term. It provides information on how variations in the concentration of reactants influence the rate.
In the given exercise, the order \(n\) was found to be \(2\). This means that the rate of reaction is proportional to the square of the concentration of \(A\). So, doubling \([A]\) will increase the rate by a factor of four, as \((2)^2 = 4\). Similarly, increasing the concentration by a factor of \(2.5\) raises the rate by \(6.25\), since \[ (2.5)^2 = 6.25 \] Orders of reaction can be whole numbers, fractions, or even zero. They reflect:
  • The pathway or mechanism of the reaction.
  • The concentration effects on reaction speed.
It is crucial to understand that the order is independent of the stoichiometric coefficients of the reactants in the balanced equation.
Rate Constant
The rate constant, \(k\), is a crucial component of the rate law. It provides the relation between the reaction rate and the reactant concentrations.
While the order of the reaction \(n\) tells us how the rate depends on concentrations, \(k\) contains essential information about the reaction conditions such as:
  • Temperature
  • Presence of a catalyst
  • The specific nature of the reactants
The units of \(k\) vary depending on the reaction order. For a second-order reaction, like in this exercise, the units will be \(M^{-1} \cdot s^{-1}\), reflecting the dependency on concentration and time.
The value of \(k\) is determined experimentally and varies with temperature according to the Arrhenius equation. This makes rate constants valuable for comparing reaction dynamics under different conditions. They are often used to calculate predicted reaction rates for given reactant concentrations.

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Most popular questions from this chapter

For an endothermic reaction, where \(\Delta \mathrm{H}\) represents the enthalpy of the reaction in \(\mathrm{kJ} / \mathrm{mol}\), the minimum value for the energy of activation will be (a) less than \(\Delta \mathrm{H}\) (b) zero (c) more than \(\Delta \mathrm{H}\) (d) equal to \(\Delta \mathrm{H}\)

Consider the chemical reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) The rate of this reaction can be expressed in terms of time derivatives of concentration of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions. (a) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (b) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] \mathrm{dt}=-3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (c) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (d) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\)

Identify the correct statements: (a) The order of an elementary reaction is equal to its molecularity (b) The order of a reaction can be zero (c) For second order reaction, order of reaction \(=2 \times\) molecularity. (d) The order of inversion of cane sugar is 2 .

For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{AB}\) this following rate data were obtained at \(300 \mathrm{~K}\). Table \(10.4\) \begin{tabular}{llll} \hline Expt & Concentration & Rate of disappearance \\ & {\([\mathrm{A}]\)} & {\(\left[\mathrm{B}_{2}\right]\)} & of \(\mathrm{B}_{2}\left(\mathrm{~mol} \mathrm{~L} \min ^{-1}\right)\) \\ \hline \(1 .\) & \(0.015\) & \(0.15\) & \(1.8 \times 10^{-2}\) \\ \(2 .\) & \(0.09\) & \(0.15\) & \(1.08 \times 10^{-2}\) \\ \(3 .\) & \(0.015\) & \(0.45\) & \(5.4 \times 10^{-2}\) \\ \hline \end{tabular} What is the rate law? (a) \(\mathrm{r}=k[\mathrm{~A}]\left[\mathrm{B}_{2}\right]\) (b) \(r=[\mathrm{A}]^{2}\left[\mathrm{~B}_{2}\right]^{1}\) (c) \(r=k[A]\left[B_{2}\right]^{2}\) (d) \(\mathrm{r}=k\left[\mathrm{~B}_{2}\right]\)

The rate constant of a reaction at temperature 200 is 10 times less than the rate constant at \(400 \mathrm{~K}\). What is the activation energy \(\left(\mathrm{E}_{\alpha}\right)\) of the reaction? \((\mathrm{R}=\) gas constant) (a) \(1842.4 \mathrm{R}\) (b) \(921.2 \mathrm{R}\) (c) \(460.0 \mathrm{R}\) (d) \(230.3 \mathrm{R}\)

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