Say I know that the reaction quotient $\left(Q = \dfrac{[B]}{[A]}\right)$ for a simple reaction $\ce{A\to B}$, at $t = 0$, $[A] = 1$ and $[B] = 0$, at some time in the future is $Q = 1$ (or any number really).

Is there a way to calculate the concentrations of A and B, or is more information required?

  • 1
    $\begingroup$ Concentration when? $\endgroup$ – JSCoder says Reinstate Monica Dec 10 '17 at 14:03
  • $\begingroup$ just for a general case - I have used the Nernst equation to calculate Q based on the potential of an actual cell (OCV measurement over time) - but am unsure how to convert Q to actual concentrations. Thanks. $\endgroup$ – Jack Dec 10 '17 at 14:07
  • $\begingroup$ Not really. For the general reaction, you would definitely need a reaction rate coefficient. Then the reaction could be zero order (say if a catalyst was being used and was saturated) or first order. Also is the backward reaction possible? If so then you'd need a backward rate coefficient too. Again the reaction could be zero or first order. Welcome to kinetics... ;-) $\endgroup$ – MaxW Dec 10 '17 at 18:22

You have everything you need.

$$\ce{A -> B}$$

From the initial condition, stoichiometry and the conservation of mass, always:

$$\ce{[A] + [B]=1}$$

At time $t$:

$$\ce{\frac{[B]}{[A]}=1\to [B]=[A]}$$

Substitute into the first equation:

$$\ce{\to [A] + [A]=1\to 2\times [A]=1\to [A]=\frac12=[B]}$$

For a more general case:

$$\ce{[A] + [B]=n}$$

At time $t$:

$$\ce{\frac{[B]}{[A]}=Q\to [B]=Q\times [A]}$$

Substituting as above, we get:


$$\ce{[B]=\frac{n\times Q}{1+Q}}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.