Question

The rate constant for the reaction NH4+(aq) + NO2−(aq) --> N2(g) + 2H2O(l) is, k = 3.0 × 10^−4 M^-1 · s^-1. Calculate the rate of the reaction if [NH4+] = 0.36 M and [NO2−] = 0.075 M. (The reaction is in first order in regards to NH4+ as well as NO2-) Answer in scientific notation

Answer 1

Answer : The rate of the reaction is, $8.1\times 10^{-6}M/s$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The balanced equations will be:

$NH_4^+(aq)+NO_2^-(aq)\rightarrow N_2(g)+2H_2O(l)$

In this reaction, $NH_4^+$ and $NO_2^-$ are the reactants.

The rate law expression for the reaction is:

$\text{Rate}=k[NH_4^+][NO_2^-]$

As we are given that:

k = rate constant = $3.0\times 10^{-4}M^{-1}s^{-1}$

$[NH_4^+]$ = concentration of $NH_4^+$ = 0.36 M

$[NO_2^-]$ = concentration of $NO_2^-$ = 0.075 M

Now put all the given values in the above expression, we get:

$\text{Rate}=(3.0\times 10^{-4}M^{-1}s^{-1})\times (0.36M)\times (0.075M)$

$\text{Rate}=8.1\times 10^{-6}M/s$

Therefore, the rate of the reaction is, $8.1\times 10^{-6}M/s$