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Question

# The rate law for the reaction NH4^+ + NO2^- $\to$ N2 + 2 H2O is given by rate = k[NH4^+ ][NO2^- ]. At 25 C, the rate constant is 3.0 × 10^-4 /M s. Calculate the rate of the reaction at this temperature if [NH4^+] = 0.26 M and [NO2^-] = 0.080 M.

Solution

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$\textbf{Knowns}$

We have $k=3.0\times10^{-4}\;(M\cdot s)^{-1}$, [$\ce{NH4^+}$] = 0.26 $M$, [$\ce{NO2^-}$] = 0.080 $M$ for the reaction

$\begin{equation*} \ce{NH4^+(aq) + NO2^-(aq) -> N2(g) + 2H2O(l)} \end{equation*}$

The rate law is given by

$\begin{equation*} \text{rate} = k[\text{NH4}^+][\ce{NO2^-}]\end{equation*}$

We need to find the reaction rate.

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