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Chapter 13: Problem 42

Why are the units of the rate constants different for reactions of different order?

Short Answer

Expert verified
Answer: The units of rate constants are different for reactions of different orders because the reaction order determines the relationship between reactant concentrations and the reaction rate. The reaction order corresponds to the exponent of the reactant concentrations in the rate law equation, which affects the units of the rate constant. As the reaction order increases, the units of the rate constant change accordingly to ensure dimensional consistency and to represent the dependency of the reaction rate on the concentrations of the reactants.

Step by step solution

01

Understanding Reaction Order

Reaction order is a term that characterizes the relationship between the concentration of reactants and the overall rate of a chemical reaction. It is defined as the exponent to which the concentration of the reactants is raised in the rate law equation. The reaction order can be determined experimentally and is usually expressed as a whole number or a fractional value.
02

Rate Law and the Rate Constant

The rate law is a mathematical expression that describes the relationship between the rate of a chemical reaction and the concentrations of the reactants. The general form of a rate law for a reaction involving reactants A and B is: Rate = k[A]^m[B]^n Here, Rate is the reaction rate, [A] and [B] are the concentrations of the reactants, k is the rate constant, and m and n are the reaction orders with respect to reactants A and B, respectively.
03

Units of Rate Constant and its Dependence on Reaction Order

To derive the units of the rate constant, we need to consider that the rate of a reaction is usually expressed in terms of concentration per unit time, usually in mol/L s or M/s. Therefore, the unit for rate constant k depends on the units of the reactant concentrations and the overall reaction order. Let's consider the overall reaction order to be p, where p = m + n. Now, we rearrange the rate law equation to make k the subject: k = Rate / ([A]^m[B]^n) The units of k can be represented as follows: k_unit = (Rate_unit) / (concentration_unit^p) For a 0th order reaction (where p = 0), the units of k are M/s. For a 1st order reaction (where p = 1), the units of k are 1/s. For a 2nd order reaction (where p = 2), the units of k are L/mol s (or M^-1 s^-1). For a 3rd order reaction (where p = 3), the units of k are L^2/mol^2 s (or M^-2 s^-1).
04

Conclusion

The units of the rate constants are different for reactions of different orders because the reaction order determines the relationship between reactant concentrations and the reaction rate. In the rate law equation, the reaction order corresponds to the exponent of the reactant concentrations; thus, affecting the units of the rate constant, which needs to ensure dimensional consistency in the equation. As the reaction order increases, the units of the rate constant change accordingly to represent the dependency of the reaction rate on the concentrations of the reactants.

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Most popular questions from this chapter

Adding NH \(_{3}\) to the stack gases at an electric power generating plant can reduce \(\mathrm{NO}_{x}\) emissions. This selective noncatalytic reduction (SNR) process depends on the reaction between \(\mathrm{NH}_{2}\) (an odd- electron compound) and NO: $$\mathrm{NH}_{2}(g)+\mathrm{NO}(g) \rightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ The following kinetic data were collected at \(1200 \mathrm{K}\) $$\begin{array}{cllc}\text { Experiment } & \left[\mathrm{NH}_{2}\right]_{0}(\mathrm{M}) & [\mathrm{NO}]_{0}(M) & \text { Rate }(M / \mathrm{s}) \\\1 & 1.00 \times 10^{-5} & 1.00 \times 10^{-5} & 0.12 \\\\\hline 2 & 2.00 \times 10^{-5} & 1.00 \times 10^{-5} & 0.24 \\\\\hline 3 & 2.00 \times 10^{-5} & 1.50 \times 10^{-5} & 0.36 \\\\\hline 4 & 2.50 \times 10^{-5} & 1.50 \times 10^{-5} & 0.45 \\\\\hline\end{array}$$ a. What is the rate law for the reaction? b. What is the value of the rate constant at \(1200 \mathrm{K} ?\)

On the basis of the frequency factors and activation energy values of the following two reactions, determine which one will have the larger rate constant at room temperature \((298 \mathrm{K})\). \(\mathrm{O}_{3}(g)+\mathrm{Cl}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)\) \(A=2.9 \times 10^{-11} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \quad E_{2}=2.16 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) \(A=2.0 \times 10^{-12} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \quad E_{\mathrm{a}}=11.6 \mathrm{kJ} / \mathrm{mol}\)

Laughing Gas Nitrous oxide ( \(\mathrm{N}_{2} \mathrm{O}\) ) is used as an anesthetic (laughing gas) and in aerosol cans to produce whipped cream. It is a potent greenhouse gas and decomposes slowly to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{N}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ If the plot of \(\ln \left[\mathrm{N}_{2} \mathrm{O}\right]\) as a function of time is linear, what is the rate law for the reaction? b. How many half-lives will it take for the concentration of \(\mathrm{N}_{2} \mathrm{O}\) to reach \(6.25 \%\) of its original concentration?

Ammonia reacts with nitrous acid to form an intermediate, ammonium nitrite (NH_NO \(_{2}\) ), which decomposes to \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}:\) \(\mathrm{NH}_{3}(g)+\mathrm{HNO}_{2}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) a. The reaction is first order in ammonia and second order in nitrous acid. What is the rate law for the reaction? What are the units on the rate constant if concentrations are expressed in molarity and time in seconds? b. The rate law for the reaction has also been written as $$ \text { Rate }=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}-\right]\left[\mathrm{HNO}_{2}\right] $$ Is this expression equivalent to the one you wrote in part \((a) ?\) c. With the data in Appendix \(4,\) calculate the value of \(\Delta H_{\text {ren }}^{\circ}\) for the overall reaction \(\Delta H_{\mathrm{f}, \mathrm{HNO}_{2}(a q)}^{\circ}=\) \(-128.9 \mathrm{kJ} / \mathrm{mol}\) d. Draw a reaction-energy profile for the process with the assumption that \(E_{\mathrm{a}}\) of the first step is lower than \(E_{\mathrm{a}}\) of the second step.

Solutions of nitrous acid (HNO \(_{2}\) ) in \(^{18} \mathrm{O}\) -labeled water undergo isotope exchange: $$\mathrm{HNO}_{2}(a q)+\mathrm{H}_{2}^{18} \mathrm{O}(\ell) \rightarrow \mathrm{HN}^{18} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)$$ a. Use the following data at \(24^{\circ} \mathrm{C}\) to determine the dependence of the reaction rate on the concentration of \(\mathrm{HNO}_{2}\) \begin{array}{cc}\text { Time (min) } & {\left[\mathrm{HNO}_{2}\right]} \\\\\hline 0 & 5.4 \times 10^{-2} \\\\\hline 20 & 1.5 \times 10^{-3} \\\\\hline 40 & 7.7 \times 10^{-4} \\\\\hline 60 & 5.2 \times 10^{-4} \\\\\hline\end{array} b. Does the reaction rate depend on the concentration of \(\mathrm{H}_{2}^{18} \mathrm{O} ?\)
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