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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.

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Knowns\textbf{Knowns}

We have k=3.0×104  (Ms)1k=3.0\times10^{-4}\;(M\cdot s)^{-1}, [NHX4X+\ce{NH4^+}] = 0.26 MM, [NOX2X\ce{NO2^-}] = 0.080 MM for the reaction

NHX4X+(aq)+NOX2X(aq)NX2(g)+2HX2O(l)\begin{equation*} \ce{NH4^+$(aq)$ + NO2^-$(aq)$ -> N2$(g)$ + 2H2O$(l)$} \end{equation*}

The rate law is given by

rate=k[NH4+][NOX2X]\begin{equation*} \text{rate} = k[\text{NH4}^+][\ce{NO2^-}]\end{equation*}

We need to find the reaction rate.

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