Question

NH4+(aq) + NO2−(aq) → N2(g) + 2H2O(l) is given by rate = k[NH4+][NO2−]. At a certain temperature, the rate constant is 4.10 × 10−4 /M·s. Calculate the rate of the reaction at that temperature if [NH4+] = 0.301 M and [NO2−] = 0.160 M.

Answer 1

Answer:

$rate = 1.97\times 10^{-5}\ M/s$

Explanation:

Given that:-

$rate = k[NH_4^+][NO_2^-]$

k = $4.10\times 10^{-4}\ /Ms$

$[NH_4^+]=0.301\ M$

$[NO_2^-]=0.160\ M$

So,

$rate = 4.10\times 10^{-4}\times 0.301\times 0.160\ M/s=1.97\times 10^{-5}\ M/s$

$1.97\times 10^{-5}\ M/s$ is the rate of the reaction at that temperature if [NH4+] = 0.301 M and [NO2−] = 0.160 M.