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How do I relate the half life to the overall rate of reaction?

I argued that from the data, doubling the partial pressure of either reactant, keeping the other constant, will half the half life.

So try t1/2 = $\frac{\ce{[1]}}{\ce{[k][PA_0][PB_0]}}$ and since pressure of A is always greater than B, so t1/2 = $\frac{\ce{[1]}}{\ce{[k'][PB_0]}}$ ?

I then wrote: $$\frac{-\mathrm{d}[PB]}{\mathrm{d}t} = k [P_B]^m $$, where k = k[PA]


Recall the meaning of the half life $t_{1/2}$: At $t_{1/2}$, a concentration (or partial pressure) is decreased to the half of its initial value.

It is crucial to realize that this represents exactly the $\mathbf{-\frac{dp_A}{dt}}$ in your rate equation!

With other words, $-\frac{dp_A}{dt} = k\cdot p_A\cdot p_B^2$ becomes $-\frac{54}{2\cdot100} = k\cdot54\cdot(1.0)^2$ for your first set of data.

Solve for $k$ and do the same for the other data sets. What is the result?

  • $\begingroup$ I'm trying to understand how you can say that the concentration at half life can be equal to the rate of reaction - it looks like you are not considering the differential? It may be that it works out for this problem, but it seems like the first step should be to integrate the proposed rate equation, then make the substitutions. $\endgroup$ – thomij Jun 8 '15 at 15:50

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