The rate law for the reaction: NH4+ + NO₂ N₂ + 2H₂O is rate =k[NH4+][NO₂7 It is found that the rate constant for this reaction is 3.05x10-4 M-1s-1. Calculate the rate of the reaction, including units, if [NH4*] is 0.265M and [NO₂] is 0.0803M. Attach File Browse Local Files

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The rate law for the reaction:

\[ \text{NH}_4^+ + \text{NO}_2^- \rightarrow \text{N}_2 + 2\text{H}_2\text{O} \]

is 

\[ \text{rate} = k[\text{NH}_4^+][\text{NO}_2^-] \]

It is found that the rate constant for this reaction is \( 3.05 \times 10^{-4} \, \text{M}^{-1}\text{s}^{-1} \). Calculate the rate of the reaction, including units, if \([\text{NH}_4^+]\) is 0.265 M and \([\text{NO}_2^-]\) is 0.0803 M.
Transcribed Image Text:The rate law for the reaction: \[ \text{NH}_4^+ + \text{NO}_2^- \rightarrow \text{N}_2 + 2\text{H}_2\text{O} \] is \[ \text{rate} = k[\text{NH}_4^+][\text{NO}_2^-] \] It is found that the rate constant for this reaction is \( 3.05 \times 10^{-4} \, \text{M}^{-1}\text{s}^{-1} \). Calculate the rate of the reaction, including units, if \([\text{NH}_4^+]\) is 0.265 M and \([\text{NO}_2^-]\) is 0.0803 M.
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