Question

question 9 (5 points)
listen
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. whats the equilibrium constant for this reaction?
a) 2.007
b) 1.417
c) 0.706
d) 0.498

Explanation:

Step1: Recall equilibrium constant formula

For a reversible reaction, the equilibrium constant \( K \) is the ratio of the forward rate constant \( k_f \) to the reverse rate constant \( k_r \), so \( K=\frac{k_f}{k_r} \).

Step2: Substitute values

Given \( k_f = 0.412\space mol/L/s \) and \( k_r = 0.827\space mol/L/s \), substitute into the formula: \( K=\frac{0.412}{0.827} \).

Step3: Calculate the result

\( \frac{0.412}{0.827}\approx0.498 \).

Answer:

D) 0.498