Question

For a different reaction, K_c = 2.95 × 10⁴ , k_f = 1.72 × 10⁵ (s)⁻¹ , and kᵣ = 5.83 (s)⁻¹ . Adding a catalyst increases the forward rate constant to 4.94 × 10⁷ (s)⁻¹ . What is the new value of the reverse reaction constant, kᵣ , after adding the catalyst? Express your answer with the appropriate units.

a. 4.94 × 10⁷ (s)⁻¹
b. 2.95 × 10⁴ (s)⁻¹
c. 5.83 × 10⁴ (s)⁻¹
d. 1.72 × 10⁵ (s)⁻¹

Answer 1

The new value of the reverse reaction constant, kr, after adding a catalyst is 5.96 × 10⁻⁴ (s)⁻¹.

Step-by-step explanation:

When a catalyst is added to a reaction, it increases the rate constant for the forward reaction. In this case, the forward rate constant increases to 4.94 × 10⁷ (s)⁻¹. The equilibrium constant, Kc, remains the same and is given as 2.95 × 10⁴. The equilibrium constant is the ratio of the forward and reverse rate constants. Therefore, to find the new value of the reverse reaction constant, kr, we can rearrange the equation Kc = kf/kr and solve for kr. rac{K_c}{k_f} = k_r and substitute the given values to calculate kr, we get, rac{2.95 imes 10⁴}{4.94 imes 10⁷} = k_r = 5.96 imes 10⁻⁴ (s)⁻¹.