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Chapter 11: Problem 85

Consider two reaction vessels, one containing A and the other containing \(\mathrm{B},\) with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$\begin{aligned} &k_{A}=4.50 \times 10^{-4} \mathrm{s}^{-1}\\\ &k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1} \end{aligned}$$how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{B}] ?\)

Short Answer

Expert verified
The time required to reach a condition where the concentration of A is 4 times the concentration of B is approximately \(340.12\, s\).

Step by step solution

01

Write the first-order reaction equations for both substances

For both substances A and B, we can write the first-order reaction equation as: \[A = A_0e^{-k_At}\] \[B = B_0e^{-k_Bt}\] Where: - \(A\) and \(B\) are the concentrations of A and B at time \(t\), - \(A_0\) and \(B_0\) are the initial concentrations of A and B, respectively, - \(k_A\) and \(k_B\) are the rate constants, - \(t\) is the time for which we need to find.
02

Set the given condition ([A] = 4[B]) and solve for t

We have the given condition that at some time 't', \([A] = 4[B]\). So, \[A_0e^{-k_At} = 4 \times B_0e^{-k_Bt}\] Since we are given that initially A and B have equal concentration (\(A_0 = B_0\)), we can rewrite the equation as: \[e^{-k_At} = 4 \times e^{-k_Bt}\] Now, we need to solve this equation for time, \(t\).
03

Solve the equation to get the value of t

Take the natural log of both sides of the equation: \[-k_At \times ln(e) = ln(4) - k_Bt \times ln(e)\] Simplify and rearrange for \(t\): \[t = \frac{ln(4)}{k_B - k_A}\] Now, substitute the given values of \(k_A\) and \(k_B\) (\(k_A = 4.50 \times 10^{-4}\, \mathrm{s^{-1}}\) and \(k_B = 3.70 \times 10^{-3}\, \mathrm{s^{-1}}\)): \[t = \frac{ln(4)}{3.70 \times 10^{-3} \mathrm{s^{-1}} - 4.50 \times 10^{-4} \mathrm{s^{-1}}}\]
04

Calculate the time for the given condition

Now, calculate the value of \(t\): \[t = \frac{ln(4)}{3.25 \times 10^{-3} \mathrm{s^{-1}}}\] \[t \approx 340.12\, s\] So, approximately \(340.12\, s\) must pass to reach a condition where the concentration of A is 4 times the concentration of B.

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Most popular questions from this chapter

The reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow\) products \((a q)\) was studied, and the following data were obtained: What is the order of the reaction with respect to A? What is the order of the reaction with respect to B? What is the value of the rate constant for the reaction?

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.

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?

Draw a rough sketch of the energy profile for each of the following cases: a. \(\Delta E=+10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=25 \mathrm{kJ} / \mathrm{mol}\) b. \(\Delta E=-10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\) c. \(\Delta E=-50 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\)

Enzymes are kinetically important for many of the complex reactions necessary for plant and animal life to exist. However, only a tiny amount of any particular enzyme is required for these complex reactions to occur. Explain.
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