Question

Given above is the concentration vs time plot for a dissociation reaction : $A \to nB$. Based on the data of the initial phase of the reaction (initial 10 min), the value of n is ____.

• A. 4
• B. 5
• C. 2
• D. 3

Hint:

The ratio of changes in concentration $\Delta[Product] / |\Delta[Reactant]|$ gives the stoichiometric coefficient of the product (assuming reactant coefficient is 1).

Answer 1

The Correct Option is A

To determine the value of \( n \) in the dissociation reaction

\[ A \rightarrow nB, \]

we analyze the concentration–time plot during the initial 10 minutes.

---

Observations from the graph:

1. The concentration of \( A \) decreases with time.
2. The concentration of \( B \) increases with time.
3. According to the reaction stoichiometry, for every mole of \( A \) consumed, \( n \) moles of \( B \) are produced.

---

Estimating concentration changes from the plot:

• Initial concentration of \( A \): \( [A]_0 = 0.05\,\text{M} \)
• Concentration of \( A \) after 10 minutes: \( [A] = 0.04\,\text{M} \)

Change in concentration of \( A \):

\[ \Delta[A] = 0.05 - 0.04 = 0.01\,\text{M} \]

• Initial concentration of \( B \): \( 0\,\text{M} \)
• Concentration of \( B \) after 10 minutes: \( 0.04\,\text{M} \)

Change in concentration of \( B \):

\[ \Delta[B] = 0.04 - 0 = 0.04\,\text{M} \]

---

Applying stoichiometry:

For the reaction \( A \rightarrow nB \):

\[ n\,\Delta[A] = \Delta[B] \]

\[ n \times 0.01 = 0.04 \]

\[ n = \frac{0.04}{0.01} = 4 \]

---

Final Answer:

\(\boxed{4}\)