Question

For the reaction, $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ $N_2O_5$ disappears at a rate of $x$ moldm$^{-3}$ s$^{-1}$ Find the rate of formation of $O_2$ ?

• A. $x$ moldm$^{-3}$ s$^{-1}$
• B. $2x$ moldm$^{-3}$ s$^{-1}$
• C. $\frac{x}{2}$ moldm$^{-3}$ s$^{-1}$
• D. $\frac{3x}{2}$ moldm$^{-3}$ s$^{-1}$

Hint:

Rate of reaction is the change in concentration divided by its stoichiometric coefficient.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question provides a balanced chemical reaction and the rate of consumption (disappearance) of one reactant ($N_2O_5$). We need to find the rate of production (formation) of a product ($O_2$).
Step 2: Key Formula or Approach:
For a general reaction \( aA + bB \rightarrow cC + dD \), the rate of reaction is given by:
\[ Rate = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt} \]
Step 3: Detailed Explanation:
The given reaction is: \( 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) \)
The relationship between the rates is:
\[ -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{1} \frac{d[O_2]}{dt} \]
The rate of disappearance of \( N_2O_5 \) is given as:
\[ -\frac{d[N_2O_5]}{dt} = x \text{ moldm}^{-3} \text{s}^{-1} \]
Substituting this into the relationship:
\[ \frac{1}{2} \times x = \frac{d[O_2]}{dt} \]
\[ \text{Rate of formation of } O_2 = \frac{x}{2} \text{ moldm}^{-3} \text{s}^{-1} \]
Step 4: Final Answer:
The rate of formation of \( O_2 \) is \( \frac{x}{2} \).