Question

An example of steady-state kinetics is the production of ¹²C in stars, which proceeds mainly via a two-step process:

Step 1: ⁴He + ⁴He ↔ ⁸Be
Step 2: ⁴He + ⁸Be → ¹²C

In this mechanism, k_-1 >> k₁ >> k₂, and k_-2 is negligible. Would you expect the amount of ⁸Be to be large or small compared to the amount of ⁴He? Justify your answer.

Answer 1

According to the given data:

The two steps of the mechanism for the production of ¹²C in stars is shown below with rate constants.

Step 1: ⁴He + ⁴He ↔ ⁸Be; k₁ (forward reaction rate constant) and k_-1 (backward reaction rate constant)
Step 2: ⁴He + ⁸Be → ¹²C, rate constant = k₂

Here, k_-1 >> k₁ >> k₂

Hence, you would expect the amount of ⁸Be to be small compared to the amount of ⁴He?

Explanation: Since the rate constant for the formation of ⁴He (k₂) is more than that for the formation of ⁸Be (k₁). Further, the rate constant for the formation of 12C from ⁴He and ⁸Be is again less than that of the former rare constants. Hence, it is sure that the amount of ⁴He is more than that of ⁸Be.​