5)

The reaction $2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ is first order in $\mathrm{H}_{2}$ and second order in $\mathrm{NO}$.
a. Write the rate law for the reaction and specify the units of the rate constant $k$ if the reaction progress is measured in concentration ( $\mathrm{mol}$ litre $^{-1}$ ).
b. Repeat part a for the case if the reaction progress is measured in partial pressure $(\mathrm{Pa})$.

Question

5)

The reaction $2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ is first order in $\mathrm{H}_{2}$ and second order in $\mathrm{NO}$.
a. Write the rate law for the reaction and specify the units of the rate constant $k$ if the reaction progress is measured in concentration ( $\mathrm{mol}$ litre $^{-1}$ ).
b. Repeat part a for the case if the reaction progress is measured in partial pressure $(\mathrm{Pa})$.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the rate law based on the given reaction orders**: Given that the reaction is first order in $\mathrm{H}_2$ and second order in $\mathrm{NO}$, the rate law can be expressed as the product of the concentration of $\mathrm{H}_2$ raised to the first power and the concentration of $\mathrm{NO}$ raised to the second power. ‖ ‖ 2 ⋮ **Express the rate law with the rate constant $k$**: The rate law for the reaction can be written as: Rate = $k[\mathrm{NO}]^2[\mathrm{H}_2]$. ‖ ‖ 3 ⋮ **Specify the units of the rate constant $k$**: The units of the rate constant can be determined by using the units of rate (mol litre$^{-1}$ s$^{-1}$) and the units of concentration (mol litre$^{-1}$). Since the rate law involves the concentration of $\mathrm{NO}$ squared and the concentration of $\mathrm{H}_2$ to the first power, the units of $k$ will be such that when multiplied by the units of concentration raised to the respective powers, the result will be the units of rate. ‖ ‖ 4 ⋮ **Calculate the units of $k$ for concentration**: The units of $k$ will be litre$^2$ mol$^{-2}$ s$^{-1}$, which is derived from the rate law and the units of rate and concentration. ‖ ‖ 5 ⋮ **Repeat the process for partial pressure**: When the reaction progress is measured in partial pressure, the rate law remains the same, but the units of $k$ will change because the units of rate will be Pa s$^{-1}$ and the units of concentration will be Pa. ‖ ‖ 6 ⋮ **Calculate the units of $k$ for partial pressure**: The units of $k$ when using partial pressure will be Pa$^{-1}$ s$^{-1}$, which is derived from the rate law and the units of rate and partial pressure. ‖ ∻1 Answer∻ ⚹ The rate law for the reaction is Rate = $k[\mathrm{NO}]^2[\mathrm{H}_2]$. The units of the rate constant $k$ when concentration is used are litre$^2$ mol$^{-2}$ s$^{-1}$. ⚹ ∻2 Answer∻ ⚹ When partial pressure is used to measure reaction progress, the units of the rate constant $k$ are Pa$^{-1}$ s$^{-1}$. ⚹ ∻Key Concept∻ ⚹ The rate law expresses the rate of a chemical reaction in terms of the concentration or partial pressure of the reactants and the specific rate constant $k$. ⚹ ∻Explanation∻ ⚹ The units of the rate constant $k$ depend on the overall order of the reaction and the units used to measure the concentration or partial pressure of the reactants. For this reaction, the units of $k$ are litre$^2$ mol$^{-2}$ s$^{-1}$ when concentration is used, and Pa$^{-1}$ s$^{-1}$ when partial pressure is used.◊What is the overall order of the reaction described in the given information?⍭ Generate me a similar question◊⚹

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