The mechanism for NH2NO2=N2O+H2O occurs in three steps:
step 1: NH2NO2=NHNO2- + H+ (forward and reverse reactions are fast)
step 2: NHNO2-=N2O+OH- (slow)
step 3: H+ + OH-=H20 (very fast)
What is the rate law for d[N2O]/dt as determined by the mechanism?
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To determine the rate law for d[N2O]/dt based on the given mechanism, we need to identify the slowest step, also known as the rate-determining step.
According to the mechanism, step 2 is the slow step: NHNO2- reacts with OH- to form N2O. Since step 2 determines the overall rate of the reaction, the rate law will be based on the concentrations of the reactants involved in this step.
The balanced equation for step 2 is: NHNO2- + OH- → N2O
Now let's determine the rate law for step 2. To do this, we'll assume that the rate of the reaction is proportional to the concentration of the reactants, raised to some powers.
Let's say the rate law for step 2 is:
rate = k[NHNO2-]^m[OH-]^n
To find the values of m and n, we can use the stoichiometry of the balanced equation. From the balanced equation, we see that the stoichiometric coefficient of NHNO2- is 1, and the stoichiometric coefficient of OH- is 1. Therefore, the rate law for step 2 can be simplified to:
rate = k[NHNO2-][OH-]
Since the concentration of OH- is not given in the mechanism, it implies that OH- is in excess and its concentration remains approximately constant throughout the reaction. Therefore, we can treat OH- as a constant, and the rate law for N2O formation in terms of [NHNO2-] only becomes:
rate = k'[NHNO2-]
Hence, the rate law for the formation of N2O (d[N2O]/dt) as determined by the given mechanism is rate = k'[NHNO2-]. Note that the rate constant (k') includes the effect of both k and the constant concentration of OH-.