Suppose reactant $\ce{A}$ decomposes in an aqueous solution to give $\ce{P}, \ce{Q}$: $$ \ce{A(aq) -> P(aq) + Q(aq)} $$

Suppose further that $[\ce{A}]_0=0.42\text{M}$, and that $[\ce{Q}]$ was measured as follows (in M):

\begin{align} t&=0 &[\ce{Q}]&=0 \\ t&=100 \quad &[\ce{Q}]&=0.101\\ t&=200 \quad &[\ce{Q}]&=0.165\\ t&=300 \quad &[\ce{Q}]&=0.206\\ t&=400 \quad &[\ce{Q}]&=0.239\\ t&=500 \quad &[\ce{Q}]&=0.263 \end{align}

Using the graph of $\ln [\ce{A}]$ and $\frac 1 {[\ce{A}]}$ vs $t$, determine if the reaction is of order $1$ or $2$.

I know how to solve these kind of problems when I'm given the concentration of $[\ce{A}]$, I just plot the graph and check which one resembles a line more, and the slope gives me the velocity constant $k$.

But here $[\ce{A}]$ is not given. I supposed it would be $[\ce{A}](t)=[\ce{A}]_0-[\ce{Q}](t)$, but I was told that was wrong. Could someone help me out?

  • $\begingroup$ I'd go with $[A](t)=[A]_0-[Q](t)$ anyway and see where it takes me. $\endgroup$ – Ivan Neretin Apr 17 '16 at 22:22
  • $\begingroup$ It had better be $\left[A\right]\!\left(t\right) = \left[A\right]_0 - \left[Q\right]\!\left(t\right)$, otherwise there's a major mass balance problem in the system! $\endgroup$ – hBy2Py Apr 17 '16 at 22:33
  • $\begingroup$ @Brian Could you elaborate? I don't understand where the mass balance problem would be. $\endgroup$ – YoTengoUnLCD Apr 17 '16 at 22:41
  • $\begingroup$ @YoTengoUnLCD See my response in chat $\endgroup$ – hBy2Py Apr 18 '16 at 2:46
  • $\begingroup$ @Brian Thank you for your answer, I was able to read it just now, as for some reason my phone didn't let me get into chat. My question was more towards as why the concentration of $P$ doesn't affect $[A]$ (my understanding of this reaction is that $A$ dissolves into $P+Q$, thus $ 1 M$ of $A$ should correspond with $1 M$ of $Q$, but rather with $1M$ of $P+Q$...). Sorry, I'm really confused. $\endgroup$ – YoTengoUnLCD Apr 18 '16 at 20:43

If the stoichiochemistry is as you present it:

$$\ce{A -> P + Q}$$

Then if at $t=0$, $[\ce{A}]=[\ce{A}]_0$ and $[\ce{Q}]=0$, then because one equivalent of $\ce{Q}$ is formed for each equivalent of $\ce{A}$ consumed, your method of determining $[\ce{A}]_t$ must be correct.

$$[\ce{A}]_t = [\ce{A}]_0 - [\ce{Q}]_t$$

Perhaps the error your instructor noted came in applying your relationship or interpreting your graph.


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