Question

NH4+(aq) + NO2-(aq) →N2(g) + 2H2O(l) Suppose rate = k[NH4+] [NO2-] The rate of this reaction is 5.4 x 10-7 mol/L.s when [ NH4+]= 0.0100 mol/L and [NO2-]=0.200 mol/L. Calculate the new rate of the reaction when [ NH4+]= 0.100mol/L and [NO2-]=0.255 mol/L.

Answer 1

The new rate of the reaction is 1.377 x 10-8 mol/L.s.

Step-by-step explanation:

To calculate the new rate of the reaction, we can use the rate equation:

rate = k[NH4+][NO2-]

Substituting the given values, the initial rate is 5.4 x 10-7 mol/L.s when [NH4+] = 0.0100 mol/L and [NO2-] = 0.200 mol/L.

Now, we can calculate the new rate using the new concentrations:

rate = k[NH4+][NO2-] = (5.4 x 10-7 mol/L.s)(0.100 mol/L)(0.255 mol/L) = 1.377 x 10-8 mol/L.s