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

`rate = 1.97* 10^(-5)\ M/s`

Step-by-step explanation:

Given that:-

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

k =
`4.10* 10^(-4)\ /Ms`

`[NH_4^+]=0.301\ M`

`[NO_2^-]=0.160\ M`

So,

`rate = 4.10* 10^(-4)* 0.301* 0.160\ M/s=1.97* 10^(-5)\ M/s`

`1.97* 10^(-5)\ M/s` is the rate of the reaction at that temperature if [NH4+] = 0.301 M and [NO2−] = 0.160 M.