Question

A reversible reaction has a forward rate constant of 0.412 mol/L/s and a reverse reaction rate constant of 0.827 mol/L/s. What's the equilibrium constant for this reaction?Question 12 options:A) 1.417B) 2.007C) 0.706D) 0.498

Answer 1

INFORMATION:

We know that:

- a reversible reaction has a forward rate constant of 0.412 mol/L/s

- the reverse reaction rate constant of the reaction is 0.827 mol/L/s

And we must find the equilibrium constant for this reaction

STEP BY STEP EXPLANATION:

To find it, we need to use that:

So, the equilibrium constant is

$K_{eq}=\frac{\text{ forward rate constant}}{\text{ reverse rate constant}}$

Now, replacing the given information in the formula:

- forward rate constant = 0.412 mol/L/s

- reverse rate constant = 0.827 mol/L/s

$K_{eq}=\frac{0.412}{0.827}=0.498$

Finally, the equilibrium constant for this reaction is 0.498

ANSWER:

D) 0.498