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Chapter 11: Problem 102

Consider a reaction of the type aA \(\longrightarrow\) products, in which the rate law is found to be rate \(=k[\mathrm{A}]^{3}\) (termolecular reactions are improbable but possible). If the first half-life of the reaction is found to be \(40 .\) s, what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: $$\text { Rate }=\frac{-d[\mathrm{A}]}{d t}=k[\mathrm{A}]^{3}$$

Short Answer

Expert verified
The time for the second half-life (T₁/₂,₂) of the termolecular reaction is equal to \(T_{1/2,2} = \frac{1}{20} [\mathrm{A}_{0}]^2 - 40\) seconds.

Step by step solution

01

Integrate the differential rate law equation

We are given the differential rate law equation: \[\frac{-d[\mathrm{A}]}{d t}=k[\mathrm{A}]^{3}\] Integrate both sides of the equation to obtain the integrated rate law equation: \[ \int \frac{1}{[\mathrm{A}]^3} \, d[\mathrm{A}] = \int k \, dt\]
02

Evaluate the integrals

To find the integrated rate law, we need to evaluate the integrals: \[- \frac{1}{2} [\mathrm{A}]^{-2} = kt + C\] Where C is the constant of integration.
03

Solve for the constant of integration

At t=0, [A]= [A]₀, the initial concentration of A. So, \[ - \frac{1}{2} [\mathrm{A}_{0}]^{-2} = C\]
04

Substitute the constant of integration back into the equation and find the time for the first half-life

We can now write the integrated rate law equation with C: \[- \frac{1}{2} [\mathrm{A}]^{-2} = kt - \frac{1}{2} [\mathrm{A}_{0}]^{-2}\] Since the time for the first half-life is given to be 40 s, at this moment [A] = 1/2 [A]₀. Replace the variables in the equation and solve for k: \[- \frac{1}{2} (1/2 [\mathrm{A}_{0}])^{-2} = 40k - \frac{1}{2} [\mathrm{A}_{0}]^{-2}\]
05

Solve for k

After solving the above equation for k, we will get: \[k = \frac{1}{120} [\mathrm{A}_{0}]^2\]
06

Find the time for the second half-life

Now, we will find the time for the second half-life. At this time, [A] = 1/4 [A]₀. Substitute the variables in the integrated rate law equation, and solve for the time: \[- \frac{1}{2} (1/4 [\mathrm{A}_{0}])^{-2} = k(T_{1/2,1}+T_{1/2,2}) - \frac{1}{2} [\mathrm{A}_{0}]^{-2}\] Where T₁/₂,₁ = 40 s (the first half-life) and T₁/₂,₂ (the second half-life) is the unknown value we want to find. Substitute the expression for k and solve for T₁/₂,₂: \[T_{1/2,2} = \frac{1}{20} [\mathrm{A}_{0}]^2 - 40\] The time for the second half-life (T₁/₂,₂) is equal to \(\frac{1}{20} [\mathrm{A}_{0}]^2 - 40\) seconds.

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Most popular questions from this chapter

Sulfuryl chloride undergoes first-order decomposition at \(320 .^{\circ} \mathrm{C}\) with a half-life of \(8.75 \mathrm{h}\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ What is the value of the rate constant, \(k\), in \(s^{-1}\) ? If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is 791 torr and the decomposition occurs in a \(1.25-\mathrm{L}\) container, how many molecules of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) remain after \(12.5 \mathrm{h} ?\)

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of the rate constant. c. The half-life for a reaction, a \(A \longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0}\) d. The half-life for a second-order reaction increases as the reaction proceeds.

Two isomers \((A \text { and } B)\) of a given compound dimerize as follows: $$\begin{aligned} &2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{A}_{2}\\\ &2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2} \end{aligned}$$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C} .\) In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} \mathrm{M}\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} \mathrm{M} .\) It was found that after each reaction had progressed for 3.00 min, \([\mathrm{A}]=3.00[\mathrm{B}] .\) In this case the rate laws are defined as $$\begin{array}{l} \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2} \\ \text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2} \end{array}$$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 min. b. Calculate the value of \(k_{2}\) c. Calculate the half-life for the experiment involving A.

The type of rate law for a reaction, either the differential rate law or the integrated rate law, is usually determined by which data is easiest to collect. Explain.

In the Haber process for the production of ammonia, $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?
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