174. A following reaction is occurring in a test tube:
𝐴 + 𝐵 ⇌ 𝐶
100 nM solution of A is incubated with an equal volume of 100 nM solution of B and left at 25°𝐶
till the equilibrium is reached. Finally, when the concentration of A is measured, it is found to be
10 nM. What is the approximate concentration of C when 2 nM solution of A is incubated with an
equal volume of 100 nM solution of B under identical conditions?
Choose the correct answer:
A. 0.95 nM
B. 0.8 nM
C. 0.1 nM
D. 10 nM

Introduction

In many biochemical and chemical systems, reactions are reversible, meaning they can proceed in both directions until an equilibrium is reached. The equilibrium concentrations of the substances involved depend on the initial concentrations and the nature of the reaction.

This article provides a step-by-step explanation of how to calculate the equilibrium concentration of a substance in a reversible reaction. We will use an example problem involving the reaction:

$$A+B\rightleftharpoons C$$

Problem Breakdown

The problem involves the following reaction:

$$A+B\rightleftharpoons C$$

1. Initially, 100 nM of A and 100 nM of B are mixed.

2. When equilibrium is reached, the concentration of A is found to be 10 nM.

This means that 90 nM of A has reacted with B to form C. Now, we need to calculate the concentration of C under new conditions, where A starts at 2 nM and B starts at 100 nM.

Step 1: Set up the initial concentrations for the first scenario.

In the first scenario, both A and B start at 100 nM.

• Initial concentration of A = 100 nM

• Initial concentration of B = 100 nM

• At equilibrium, concentration of A = 10 nM

This means that 90 nM of A has reacted. Because the reaction is balanced, the amount of B that reacted is also 90 nM (since 1 mole of A reacts with 1 mole of B).

So, the equilibrium concentrations are:

• A = 10 nM

• B = 10 nM (since 100 nM – 90 nM = 10 nM)

• C = 90 nM

Step 2: Set up the initial concentrations for the second scenario.

Now, for the second scenario:

• Initial concentration of A = 2 nM

• Initial concentration of B = 100 nM

In this case, because the concentration of A is much smaller, the amount of A that reacts will be limited by its initial concentration. If A reacts in a similar fashion as before (where 1 mole of A reacts with 1 mole of B), then at equilibrium:

• A will be reduced by a certain amount, but since we are starting with only 2 nM of A, we expect that A will be mostly used up.

• The concentration of C will increase by the same amount that A is consumed.

Step 3: Approximate concentration of C.

Since the initial concentration of A is 2 nM, and assuming that the reaction proceeds similarly as before, we would expect the concentration of C to increase by approximately 2 nM (the amount of A consumed).

Thus, the concentration of C in this scenario will be approximately 0.95 nM (since 2 nM of A reacts, and the reaction creates an equivalent amount of C).

Conclusion

The approximate concentration of C in the second scenario will be:

A. 0.95 nM

This is because the reaction proceeds with the available A, and at equilibrium, the concentration of C will be approximately the same as the amount of A that reacts, which is 2 nM.