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

Rate of Reaction = 6.24 x 10–6 M/s

Step-by-step explanation:

Rate of reaction = k[NH4+][NO2–]

Concentration of [NH4+] = 0.26 M

Concentration of [NO2–] = 0.080 M

k= 3.0 × 10–4/ M · s

Rate of Reaction = (3.0 × 10–4/ M · s)( 0.26 M)(0.080 M)

Answer 2

The rate of the reaction is
`6.24* 10^(-6) M/s`.

Step-by-step 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* 10^(-4) M^(-1) s^(-1)`

`R= k[NH_(4)^+][NO_(2)^-]`

`R=3.0* 10^(-4) M^(-1) s^(-1)* 0.26 M* 0.080 M`

`R=6.24* 10^(-6) M/s`

The rate of the reaction is
`6.24* 10^(-6) M/s`.