# How to calculate concentration from reaction quotient and initial concentration?

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?

• Concentration when? – JSCoder says Reinstate Monica Dec 10 '17 at 14:03
• 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. – Jack Dec 10 '17 at 14:07
• 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... ;-) – 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:

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

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