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Chapter 12: Problem 17

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

Short Answer

Expert verified
The order of the reactant that causes the initial rate to double when its concentration is quadrupled is 1. If a reactant has an order of -1, when its concentration increases by a factor of two, the initial rate of the reaction is reduced by half.

Step by step solution

01

Understand reaction order and the rate law

The reaction order indicates the relationship between the concentration of a reactant and the rate of the reaction. The rate of a reaction can be represented by a rate law, which generally takes the form: Rate = k[A]^m[B]^n where Rate is the initial rate of the reaction, k is the rate constant, [A] and [B] represent the concentrations of the reactants, and m and n are their respective orders. The overall order of the reaction is the sum of m and n.
02

Determine the order of the reactant

Given that the initial rate of the reaction doubles when the concentration of one reactant is quadrupled, let's call this reactant A. We can set up an equation representing the rate of reaction: Rate₁ = k[A₁]^m When the concentration of reactant A is quadrupled, the new rate of reaction (Rate₂) can be written as: Rate₂ = k[4A₁]^m Since the rate is doubled when the concentration is quadrupled: 2 * Rate₁ = k[4A₁]^m Now we can divide this equation by the initial rate equation (Rate₁) to solve for m: 2 = (k[4A₁]^m) / (k[A₁]^m) By canceling out k and A₁, we get: 2 = 4^m Taking the logarithm (base 2) of both sides: m = log₂(2) = 1 So the order of reactant A is equal to 1.
03

Understand the case where the reactant order is -1

Now, let's analyze the second part of the question, where a reactant has an order of -1. Suppose this reactant is B, and we want to know the effect on the initial rate when the concentration of B increases by a factor of two. We write the initial rate equation: Rate₁ = k[B₁]^{-1} When the concentration of reactant B doubles: Rate₂ = k[2B₁]^{-1}
04

Calculate the new rate for the -1 order reactant

Now, let's divide the Rate₂ equation by the Rate₁ equation to find the effect on the initial rate: (Rate₂) / (Rate₁) = (k[2B₁]^{-1}) / (k[B₁]^{-1}) By canceling out k and simplifying: (Rate₂) / (Rate₁) = 1/2 Thus, when the concentration of a reactant with an order of -1 doubles, the initial rate of the reaction is reduced by half.

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Most popular questions from this chapter

Sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) decomposes to sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) and chlorine \(\left(\mathrm{Cl}_{2}\right)\) by reaction in the gas phase. The following pressure data were obtained when a sample containing \(5.00 \times 10^{-2}\) mol sulfuryl chloride was heated to \(600 . \mathrm{K}\) in a \(5.00 \times 10^{-1}-\mathrm{L}\) container. Defining the rate as \(-\frac{\Delta\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]}{\Delta t}\), a. determine the value of the rate constant for the decomposition of sulfuryl chloride at \(600 . \mathrm{K}\). b. what is the half-life of the reaction? c. what fraction of the sulfuryl chloride remains after \(20.0 \mathrm{~h}\) ?

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A certain reaction has the following general form: \(\mathrm{aA} \longrightarrow \mathrm{bB}\) At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} \mathrm{M}\), concentration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{~min}^{-1}\). a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of \(\mathrm{A}\) to decrease to \(2.50 \times 10^{-3} M ?\)

Upon dissolving \(\operatorname{InCl}(s)\) in \(\mathrm{HCl}, \operatorname{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$ \operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ This disproportionation follows first-order kinetics with a half-life of \(667 \mathrm{~s}\). What is the concentration of \(\operatorname{In}^{+}(a q)\) after \(1.25 \mathrm{~h}\) if the initial solution of \(\mathrm{In}^{+}(a q)\) was prepared by dissolving \(2.38 \mathrm{~g} \operatorname{InCl}(s)\) in dilute \(\mathrm{HCl}\) to make \(5.00 \times 10^{2} \mathrm{~mL}\) of solution? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{~h}\) ?

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in \(\mathrm{A}\) and to have a rate constant of \(5.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\). An experiment was run at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.0 \times 10^{-3} M\) a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{~s}\) has elapsed assuming \([\mathrm{B}]_{0}=0\).
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