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The rate law for the reaction NH4(aq)+ + NO2(aq)- → H2O(l) + NO2(g) is rate = k [NH4+][NO2-], and the value of k is 2.7*10-4 M-1s-1. What is the rate of reaction when [NH4+] = 0.050 M and [NO2-] = 0.25 M?

Respuesta :

Blitztiger
We are given the rate law, so we can substitute the given values for the rate constant and the concentrations of the reactants to solve for the rate of reaction. Since rate = k [NH4+][NO2-]:
rate = (2.7 x 10^-4 / M-s)(0.050 M)(0.25 M) = 3.375 x 10^-6 M/s

3.4*10^6 Ms-1 You have successfully calculated the rate at the given conditions.

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