What are the units of the rate constant for a thirdorder reaction?

Short Answer

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The units of the rate constant for a third order reaction are (liters squared per mole squared) per time.

Step by step solution

01

Understand the Order of Reaction

For a chemical reaction, the rate law is given by: Rate = k[A]^m[B]^n where A and B are reactants, m and n are orders with respect to reactants A and B, and k is the rate constant. The overall order of the reaction is the sum of the individual orders, that is m + n.
02

Relate Order of Reaction to Units of Rate Constant

For a third order reaction, the rate of the reaction is directly proportional to the cube of the concentration of the reactants (total order = 3). The overall reaction rate is given by: Rate = k[A]^3 Considering the units: - Rate of reaction is typically in moles / liter / time - Concentration [A] is typically in moles / liter Therefore, to balance the units on both sides of the equation, k, the rate constant, must be in (liter^2 / mole^2) / time.
03

State the Units of Rate Constant for a Third Order Reaction

In conclusion, for a third order reaction, the units of the rate constant (k) are (liters squared per mole squared) per time. This is because the rate constant must compensate for the units of rate concentration to the power of the reaction order in the rate law expression.

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Most popular questions from this chapter

Consider the reaction $$ \mathrm{X}+\mathrm{Y} \longrightarrow \mathrm{Z} $$ From the following data, obtained at \(360 \mathrm{~K}\), (a) determine the order of the reaction, and (b) determine the initial rate of disappearance of \(\mathrm{X}\) when the concentration of \(\mathrm{X}\) is \(0.30 \mathrm{M}\) and that of Y is \(0.40 \mathrm{M}\). $$ \begin{array}{ccc} \hline \text { Initial Rate of } & & \\ \text { Disappearance of } \mathbf{X}(\boldsymbol{M} / \mathbf{s}) & {[\mathrm{X}](M)} & {[\mathrm{Y}](M)} \\ \hline 0.053 & 0.10 & 0.50 \\ 0.127 & 0.20 & 0.30 \\ 1.02 & 0.40 & 0.60 \\ 0.254 & 0.20 & 0.60 \\ 0.509 & 0.40 & 0.30 \\ \hline \end{array} $$

To prevent brain damage, a drastic medical procedure is to lower the body temperature of someone who has suffered cardiac arrest. What is the physiochemical basis for this treatment?

Determine the overall orders of the reactions to which the following rate laws apply: (a) rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\), (b) rate \(=k,(\mathrm{c})\) rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{\frac{1}{2}}\) (d) rate \(=\) \(k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)

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Many reactions involving heterogeneous catalysts are zero order; that is, rate \(=k .\) An example is the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) over tungsten (W): $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ It is found that the reaction is independent of \(\left[\mathrm{PH}_{3}\right]\) as long as phosphine's pressure is sufficiently high \((\geq 1 \mathrm{~atm}) .\) Explain.

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