Question

For the elementary reaction  At time 𝑡 = 0, [𝐀] = 𝐴₀ and [𝐁]=[𝐂]=0. At a later time 𝑡, the value of \(\frac{[B]}{[C]}\) is ______. (round off to the nearest integer)

Answer 1

Correct Answer: 2

Consider the reaction: C⇌A⇌B with rate constants \( k_1 \) and \( k_2 \), where \( k_1 = 2k_2 \).

Initially at \( t = 0 \), \([A] = A_0\) and \([B] = [C] = 0\). 

Assumptions: The reaction reaches an equilibrium state at a later time \( t \), with concentrations \([A]\), \([B]\), and \([C]\).

As the system reaches equilibrium, apply the steady-state approximation for \( \text{[A]} \), and use the given relation \( k_1 = 2k_2 \).

At equilibrium:

1. The forward rate, \( k_1[A] \), is equal to the backward rate, \( k_2[B] \), hence: \( k_1[A] = k_2[B] \).
2. Since \( k_1 = 2k_2 \), it follows \( 2k_2[A] = k_2[B] \), thus: \([B] = 2[A]\).
3. By the reaction stoichiometry, \(([A])_0 - [A] - [B] - [C] = 0:\) \(([A])= A_0 - [B] - [C]\).
4. Given at equilibrium \([B] = 2[A]\) and \([C] = A_0 - [A] - [B]\), substitute \([B] = 2[A]\) in the equation, then: \([C] = A_0 - [A] - 2[A] = A_0 - 3[A]\).
5. To find \(\frac{[B]}{[C]}\): \(\frac{[B]}{[C]} = \frac{2[A]}{A_0 - 3[A]}\).
6. Solving the equilibrium condition with simplified concentrations: Assume \([B] = x\) for ease, then \([C] = A_0 - 3x/2\), and verify: \(\frac{x}{A_0 - \frac{3x}{2}} = 2\) implies: \(2(A_0 - \frac{3x}{2}) = x\).
7. Solve for \( x \): \(2A_0 - 3x = x\) becomes \(2A_0 = 4x\). Hence, \(x = \frac{A_0}{2}\).
8. Substitute back: \(\frac{A_0}{2}\) for \([B]\) and confirm the ratio: \(\frac{[B]}{[C]} = \frac{A_0/2}{A_0/2} = 2\).

The final ratio of \(\frac{[B]}{[C]}\) is 2, confirming it is within the given range (2,2).