What the possible rate laws are for the decomposition of nitramide by the following process:

$$ \begin{align} \ce{O2NNH2(aq) &<=>[$k_1$][$k_{-1}$] O2NNH-(aq) + H+(aq)} &\quad &\text{(fast equilibrium)} \tag{1}\\ \ce{O2NNH-(aq) &<=>[$k_2$] N2O(g) + OH-(aq)} &\quad &\text{(slow)} \tag{2}\\ \ce{H+(aq) + OH-(aq)&<=>[$k_3$] H2O(l)} &\quad &\text{(fast)} \tag{3}\\ \end{align} $$

(A) $\text{rate} = k[\ce{O2NNH-}]$
(B) $\text{rate} = k[\ce{O2NNH2}]/[\ce{H+}]$
(C) $\text{rate} = k[\ce{O2NNH2}][\ce{H+}]$
(D) Both A and B

I selected A, but the answer is actually D. The answer key gives the following information:

At the first glance, you may think that Option A is the best answer due to the information available in the question stem in which details the way a rate of reaction is written. However, option B also deserves some attention. It is important to understand that $K_\mathrm{eq}$ is the equilibrium constant for step 1, as this is the fast equilibrium step

$$K_\mathrm{eq} = \frac{[\ce{O2NNH-}][\ce{H+}]}{[\ce{O2NNH2}]}$$

Rearranging for $[\ce{O2NNH-}]$, and substituting it into the rate law gives an equally valid expression which leads to Option B also being correct. The best answer is therefore Option D.

I am confused about how this can be so.

Firstly, why can we just sub the equilibrium equation in like this, and what happens to the Ksp term, it seems to just disappear.

Secondly, doesn't this mean that the overall order of the reaction is inconsistent between option A and B (i.e. in A the overall order is 1, but in B it is 1-1=0).

Thirdly, conceptually I am confused about why the concentration of the product should change the rate of the forward reaction, this does not really make sense to me, why should the reaction proceed less quickly just because there is more product around? Further, if it is the case that the product changes the speed of the forward reaction, then why don't all rate equations feature the product?

I am left very confused, perhaps someone can help me out?

  • $\begingroup$ > (A)$ \text{rate} =K_2[\ce{O2NNH-}]$ $\endgroup$ Commented Mar 1, 2019 at 6:56
  • $\begingroup$ > (B)$ \text{rate} =\frac{ k_1K2}{K-1}\cdot{\frac{[\ce{O2NNH2}]}{[\ce{H+}]}}$ $\endgroup$ Commented Mar 1, 2019 at 6:59

2 Answers 2


The $\ce{O2NNH^-}$ is formed very rapidly but reacts slowly in the second step. The rate limiting step is therefore loss of $\ce{O2NNH^-}$ in which the back reaction with $\ce{N2O}$ is considered (in the question) to be so slow as to be insignificant when writing the rate expression. Thus the rate involves just $\ce{O2NNH^-}$, however, as the fist reaction is a fast equilibrium substituting for $\ce{O2NNH^-}$ with this equilibrium is valid, provided the $k's$ are marked as being different, which they are not, and in the second reaction $k$ should be $k\times K_{eq_1} $.


$$\text{ Production rate of}~~ \ce{O2NNH-} = \text{ Consumption rate of}~~ \ce{\ce{O2NNH-}}$$ $$k_\mathrm {1}\ce{[\ce{O2NNH2}]}= k_\mathrm{-1}[\ce{\ce{O2NNH-}}][\ce{H+}] + k_\mathrm {2} \ce{[\ce{O2NNH-}] }$$ But $k_2 << k_{-1}$ because the second step is the slow step, so the term $k_2\ce{ [\ce{O2NNH-}]}$ can be neglected at the far right of the last equation. $$k_1\ce{[\ce{O2NNH2}]}= k_\mathrm{-1}\ce{ [\ce{O2NNH-}][\ce{H+}] }$$ $$[\ce{\ce{O2NNH-}}] =\frac{ k_1 [\ce{O2NNH2}]} { k_\mathrm{-1}[\ce{H+}]}$$ The value of$ [\ce{\ce{O2NNH-}}]$ from the last equation can now be substituted in the reaction rate equation which represents the slow step.

$$\begin{array}{ } \text{Rate} = k_\mathrm{2}[\ce{O2NNH-}] \\ ~~~~~~~~= \frac{k_\mathrm{1}k_\mathrm{2}[\ce{O2NNH2}]}{k_\mathrm{-1}[\ce{H+}]}\\ ~~~~~~~~ =\frac{k[\ce{O2NNH2}]}{[\ce{H+}]}\end{array}$$

In it we see that $k$ expresses the value of the fractional $\frac{k_\mathrm{1}k_\mathrm{2}}{k_\mathrm{-1}}$


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