Solution Found!
The rate for law for the reaction is given by rate = . At
Chapter 13, Problem 13P(choose chapter or problem)
The rate for law for the reaction
\(\mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)
is given by \(\text { rate }=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}^{-}\right]\). At 25°C, the rate constant is \(3.0 \times 10^{-4} / \mathrm{M} \cdot \mathrm{s}\). Calculate the rate of the reaction at this temperature if \(\left[N H_{4}^{+}\right]\) = 0.26 M and \(\left[\mathrm{NO}_{2}^{-}\right]\) = 0.080 M.
Questions & Answers
QUESTION:
The rate for law for the reaction
\(\mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)
is given by \(\text { rate }=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}^{-}\right]\). At 25°C, the rate constant is \(3.0 \times 10^{-4} / \mathrm{M} \cdot \mathrm{s}\). Calculate the rate of the reaction at this temperature if \(\left[N H_{4}^{+}\right]\) = 0.26 M and \(\left[\mathrm{NO}_{2}^{-}\right]\) = 0.080 M.
ANSWER:
Step 1 of 2
The chemical reaction can be written as follows.
The rate of the reaction can be written as shown below.
It is given that,
Rate constant, 
