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Chapter 11: Problem 96

The following reaction is second-order in A and first-order in B. $$\mathrm{A}+\mathrm{B} \longrightarrow \text { products }$$ (a) Write the rate expression. (b) Consider the following one-liter vessel in which each square represents a mole of \(\mathrm{A}\) and each circle represents a mole of \(\mathrm{B}\). What is the rate of the reaction in terms of \(k ?\) (c) Assuming the same rate and \(k\) as (b), fill the similar one-liter vessel shown in the figure with an appropriate number of circles (representing B).

Short Answer

Expert verified
Additionally, how many moles of B are needed in a vessel with 1 mole of A to maintain the same rate? Answer: The rate of the reaction with 2 moles of A and 4 moles of B is 16k. To maintain the same rate with 1 mole of A, you would need 16 moles of B.

Step by step solution

01

(a) Writing the rate expression

The given reaction is second-order in A and first-order in B. Therefore, the rate expression can be written as: $$\text{rate} = k[\mathrm{A}]^2[\mathrm{B}]$$
02

(b) Finding the rate of the reaction in terms of k

In the given one-liter vessel, we have 2 moles of A (squares) and 4 moles of B (circles). Using the rate expression, we can find the rate in terms of k: $$\text{rate} = k[\mathrm{A}]^2[\mathrm{B}] = k(2\,\text{mol})^2(4\,\text{mol}) = 16k$$ So, the rate of the reaction is \(16k\).
03

(c) Filling the one-liter vessel with an appropriate number of circles (B) for the same rate

Since we want to maintain the same rate as in part (b) and we have the same rate constant, k, we can set up an equation using the rate expression: $$16k = k[\mathrm{A}]^2[\mathrm{B}]$$ In the new scenario, there are 1 mole of A (square) in the one-liter vessel. So, you need to find how many moles of B (circles) are required to keep the same rate: $$16k = k(1\,\text{mol})^2[\mathrm{B}]$$ $$16k = k[\mathrm{B}]$$ Now, solve for the concentration of B: $$[\mathrm{B}] = \frac{16k}{k}$$ $$[\mathrm{B}] = 16\,\text{mol}$$ Therefore, in order to maintain the same rate as in part (b), you need 16 circles (moles of B) in the one-liter vessel in the new scenario.

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Most popular questions from this chapter

The decomposition of \(\mathrm{A}\) at \(85^{\circ} \mathrm{C}\) is a zero-order reaction. It takes 35 minutes to decompose \(37 \%\) of an initial mass of \(282 \mathrm{mg}\). (a) What is \(k\) at \(85^{\circ} \mathrm{C}\) ? (b) What is the half-life of \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C}\) ? (c) What is the rate of decomposition for \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C} ?\) (d) If one starts with \(464 \mathrm{mg}\), what is the rate of its decomposition at \(85^{\circ} \mathrm{C} ?\)

Copper-64 is one of the metals used to study brain activity. Its decay constant is \(0.0546 \mathrm{~h}^{-1}\). If a solution containing \(5.00 \mathrm{mg}\) of \(\mathrm{Cu}-64\) is used, how many milligrams of Cu-64 remain after eight hours?

The first-order rate constant for the decomposition of a certain drug at \(25^{\circ} \mathrm{C}\) is \(0.215 \mathrm{month}^{-1}\) (a) If \(10.0 \mathrm{~g}\) of the drug is stored at \(25^{\circ} \mathrm{C}\) for one year, how many grams of the drug will remain at the end of the year? (b) What is the half-life of the drug? (c) How long will it take to decompose \(65 \%\) of the drug?

At low temperatures, the rate law for the reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)$$ is as follows: rate \(=\) constant \(\times\left[\mathrm{NO}_{2}\right]^{2}\). Which of the following mechanisms is consistent with the rate law? (a) \(\mathrm{CO}+\mathrm{NO}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}\) (b) \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4} \quad\) (fast) \(\mathrm{N}_{2} \mathrm{O}_{4}+2 \mathrm{CO} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{NO} \quad\) (slow) (c) \(2 \mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\) NO \(\quad\) (slow) \(\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2} \quad\) (fast) (d) \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2} \quad\) (slow) \(\mathrm{O}_{2}+2 \mathrm{CO} \longrightarrow 2 \mathrm{CO}_{2} \quad\) (fast)

For the reaction $$2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ the experimental rate expression is rate \(=k[\mathrm{NO}]^{2} \times\left[\mathrm{H}_{2}\right] .\) The following mechanism is proposed: $$\begin{array}{cc}2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} & \text { (fast) } \\ \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2} \mathrm{O} & \text { (slow) } \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array}$$ Is this mechanism consistent with the rate expression?
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