Question

The equilibrium constant for the reaction $\mathit{Q}\mathbf{\to }\mathit{R}$is 25.

1. If $\mathbf{50}\mathit{\mu }\mathit{M}$of Q is mixed with $\mathbf{50}\mathit{\mu }\mathit{M}$of R, which way will the reaction proceed to generate more Q or more R?
2. Calculate the equilibrium concentrations of Q and R.

Short answer

In the solution (a), equal concentration of $\left[\text{Q}\right]$and$\left[\text{R}\right]$ is mixed, the reaction will proceed to produce more R molecules.

In the solution (b), the equilibrium concentration of$\left[Q\right]\text{= 3.85μM}$ and $\left[\text{R}\right]\text{= 96.15μM}$.

Step-by-step solution

Given data

$\begin{array}{l}\left[Q\right]=3.85\mu M\\ \left[\mathrm{ℝ}\right]=96.15\mu M\end{array}$

${\text{K}}_{\text{eq}}$for the given reaction$\text{Q}\to \text{R}$ is 25.

Then,

$K_{eq}=\frac{\left[R\right]}{\left[Q\right]}=25$

In solution (a), the reaction will proceed to generate more R.

$K_{eq}=\frac{\left[R\right]}{\left[Q\right]}=25$

Therefore,

$\left[R\right]=25$times of$\left[Q\right]$

At equilibrium, R molecules are 25 times more than the Q molecules.

However, when equal concentration of $\left[\text{Q}\right]$and$\left[\text{R}\right]$ are mixed, the reaction will proceed forward and more Q molecules are converted to R molecules.

In solution (b), calculation of equilibrium concentration of Q and R.

Let us assume that the amount of Q converted to R is x.

Then,

$\begin{array}{rcl}& & \left[\text{R}\right]\text{= 50 + x}\\ & & \left[\text{Q}\right]\text{= 50 - x}\\ & & \text{Calculating the value of x,}\\ & & {\text{K}}_{\text{eq}}\text{=}\frac{\left[\text{R}\right]}{\left[\text{Q}\right]}\\ & & \text{25 =}\frac{\text{50 + x}}{\text{50 - x}}\\ & & \text{50 + x = 25}\left(\text{50 - x}\right)\\ & & \text{x = 46.15}\\ & & \end{array}$

Now, calculating equilibrium constants,

$\left[\text{Q}\right]\text{= 50 - 46.15 = 3.85μM}$

Similarly,

$\left[\text{R}\right]\text{= 50 + 46.15 = 96.15μM}$

Hence, the equilibrium constant of [Q] is 3.85µM and [R] is 96.15 µM.