Question

a reversible reaction has an equilibrium constant of 0.75. if the forward rate constant is 0.62 mol/l/s, whats the reverse reaction rate constant? a) 0.910 b) 0.827 c) 1.210 d) 1.010

Explanation:

Step1: Recall the relationship between equilibrium constant and rate constants

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

Step2: Rearrange the formula to solve for \( k_r \)

From \( K = \frac{k_f}{k_r} \), we can rearrange it to \( k_r=\frac{k_f}{K} \).

Step3: Substitute the given values

We know that \( k_f = 0.62 \, \text{mol/L/s} \) and \( K = 0.75 \). Substituting these values into the formula for \( k_r \), we get \( k_r=\frac{0.62}{0.75} \).

Step4: Calculate the value of \( k_r \)

\( \frac{0.62}{0.75}\approx0.827 \) (rounded to three decimal places).

Answer:

B) 0.827