Question

3. The rate law for the reaction NH4+(aq) + NO2–(aq) → N2(g) + 2H2O(l) is given by rate = k[NH4+][NO2–]. At 25ºC, 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. (5 points)

Answer 1

Answer:

The rate of the reaction is $6.24\times 10^{-6} M/s$.

Explanation:

$NH4^+(aq) + NO_2^-(aq)\rightarrow N_2(g) + 2H_2O(l)$

Concentration of $[NH_4^{+}]=0.26 M$

Concentration of $[NO_2^{-}]=0.080 M$

Rate constant of the reaction = k= $3.0\times 10^{-4} M^{-1} s^{-1}$

$R= k[NH_{4}^+][NO_{2}^-]$

$R=3.0\times 10^{-4} M^{-1} s^{-1}\times 0.26 M\times 0.080 M$

$R=6.24\times 10^{-6} M/s$

The rate of the reaction is $6.24\times 10^{-6} M/s$.