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Chapter 14: Problem 33

The reaction \(A+B \longrightarrow C+D\) is second order in \(A\) and zero order in B. The value of \(k\) is \(0.0103 \mathrm{M}^{-1} \mathrm{min}^{-1}.\) What is the rate of this reaction when \([\mathrm{A}]=0.116 \mathrm{M}\) and \([\mathrm{B}]=3.83 \mathrm{M} ?\)

Short Answer

Expert verified
The rate of reaction when \(\displaystyle [A]=0.116 M\) and \(\displaystyle [B]=3.83 M\) is \(\displaystyle 0.139 M min^{-1}\)

Step by step solution

01

Write out the rate law for the reaction

Using the information about the order of the reaction with respect to each reactant, we can write out the rate law for the reaction. For a reaction \(\displaystyle jA+kB \longrightarrow\), the rate law is usually \(\displaystyle Rate=k[A]^{j}[B]^{k}\) where \(\displaystyle k\) is the rate constant, and the exponents reflect the order of the reaction with respect to each reactant. Here, since the reaction is second order in \(\displaystyle A\) and zero order in \(\displaystyle B\), the rate law becomes \(\displaystyle Rate = k[A]^{2}[B]^{0}\)
02

Simplify the rate law expression

Since any number or quantity raised to the power of zero equals one, we can simplify the expression given above by getting rid of \(\displaystyle [B]^{0}\). This leaves us with \(\displaystyle Rate = k[A]^{2}\)
03

Substitute given values into the rate law

Substitute the given values for \(\displaystyle k\), \(\displaystyle [A]\), and \(\displaystyle [B]\) into the rate law. The rate constant \(\displaystyle k\) given is \(\displaystyle 0.0103 \mathrm{M}^{-1} \mathrm{min}^{-1}\) and the concentrations of \(\displaystyle A\) and \(\displaystyle B\) are \(\displaystyle 0.116 \mathrm{M}\) and \(\displaystyle 3.83 \mathrm{M}\), respectively. Remember that since we found the rate law for this reaction is \(\displaystyle Rate = k[A]^{2}\), we no longer need the concentration of \(\displaystyle B\). Therefore, substituting the given values gives us \(\displaystyle Rate = 0.0103 M^{-1} min^{-1} * (0.116 M)^{2}\)
04

Solve for the Rate

To get \(\displaystyle Rate\), we proceed to simplify \(\displaystyle 0.0103 M^{-1} min^{-1} * (0.116 M)^{2}\). This results in \(\displaystyle Rate = 0.139 M min^{-1}\). So, the rate of reaction when \(\displaystyle [A]=0.116 M\) is \(\displaystyle 0.139 M min^{-1}\)

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Most popular questions from this chapter

According to collision theory, chemical reactions occur through molecular collisions. A unimolecular elementary process in a reaction mechanism involves dissociation of a single molecule. How can these two ideas be compatible? Explain.

The reaction \(A \longrightarrow\) products is second order. The initial rate of decomposition of \(A\) when \([\mathrm{A}]_{0}=0.50 \mathrm{M}\) is \((\mathrm{a})\) the same as the initial rate for any other value of \([\mathrm{A}]_{0} ;\) (b) half as great as when \([\mathrm{A}]_{0}=1.00 \mathrm{M} ;(\mathrm{c})\) five times as great as when \([\mathrm{A}]_{0}=[\mathrm{A}]_{0}=0.25 \mathrm{M}.\)

The reaction \(A \longrightarrow\) products is first order in A. Initially, \([\mathrm{A}]=0.800 \mathrm{M}\) and after 54min, \([\mathrm{A}]=0.100 \mathrm{M}.\) (a) At what time is \([\mathrm{A}]=0.025 \mathrm{M} ?\) (b) What is the rate of reaction when \([\mathrm{A}]=0.025 \mathrm{M} ?\)

In your own words, define or explain the following terms or symbols: (a) \([\mathrm{A}]_{0} ;\) (b) \(\dot{k} ;\) (c) \(t_{1 / 2} ;\) (d) zeroorder reaction; (e) catalyst.

For the reaction \(A+B \longrightarrow C+D\), the following initial rates of reaction were found. What is the rate law for this reaction? $$\begin{array}{llll} \hline & & & \text { Initial Rate, } \\ \text { Expt } & \text { [A], M } & \text { [B], M } & \text { M min }^{-1} \\\ \hline 1 & 0.50 & 1.50 & 4.2 \times 10^{-3} \\ 2 & 1.50 & 1.50 & 1.3 \times 10^{-2} \\ 3 & 3.00 & 3.00 & 5.2 \times 10^{-2} \\ \hline \end{array}$$
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