Question

For a given reaction $A \rightarrow nB$, a graph is given between concentration and time. Find value of n for above reaction, based on the information given in graph for $10 \text{ min}$.

• A. 3
• B. 2
• C. 1
• D. 4

Hint:

The stoichiometric coefficient $n$ for a product B in reaction $A \rightarrow nB$ is simply the ratio of the change in concentration of B to the change in concentration of A over the same time interval.

Answer 1

The Correct Option is A

To find the value of \(n\) for the reaction \(A \rightarrow nB\), we analyze the graph provided for the reaction over a period of 10 minutes. 

Step-by-Step Solution:

From the graph, we observe the initial concentration of A at 10:00 AM is 0.5 M and at 10:10 AM it reduces to 0.3 M.

The change in concentration of A over 10 minutes:

\(\Delta [A] = 0.5 \, \text{M} - 0.3 \, \text{M} = 0.2 \, \text{M}\)

From the graph, we also observe that the concentration of B increases to 0.6 M in the same time period.

The change in concentration of B over 10 minutes:

\(\Delta [B] = 0.6 \, \text{M}\)

The stoichiometry of the reaction gives:

\(- \frac{\Delta [A]}{\Delta t} = n \cdot \frac{\Delta [B]}{\Delta t}\)

Substituting the values, we get:

\(0.2 = n \times 0.6\)

Solving for \(n\):

\(n = \frac{0.2}{0.2} = 3\)

Conclusion:

The value of \(n\) for the reaction \(A \rightarrow nB\) is 3, matching the correct option given.