Question

For the reaction 2SO$_2$ + O$_2$ \(\rightleftharpoons \) 2SO$_3$, the rate of disappearance of O$_2$ is \( 2 \times 10^{-4} \) mol L$^{-1$ s$^{-1}$. 
The rate of appearance of SO$_3$ is:
 

• A. \( 2 \times 10^{-4} \) mol L$^{-1}$ s$^{-1}$
• B. \( 4 \times 10^{-4} \) mol L$^{-1}$ s$^{-1}$
• C. \( 1 \times 10^{-1} \) mol L$^{-1}$ s$^{-1}$
• D. \( 6 \times 10^{-4} \) mol L$^{-1}$ s$^{-1}$

Hint:

In rate calculations, use the balanced chemical equation to determine the relationship between reactant disappearance and product formation.

Answer 1

The Correct Option is B

Step 1: Understanding the Reaction Stoichiometry
The given reaction is: \[ 2SO_2 + O_2 \rightleftharpoons 2SO_3 \] Step 2: Relating the Rate of Disappearance and Appearance
The rate of disappearance of O$_2$ is related to the rate of appearance of SO$_3$ using stoichiometry: \[ \frac{-d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt} \] Step 3: Substituting Values and Solving
Given: \[ \frac{-d[O_2]}{dt} = 2 \times 10^{-4} { mol L}^{-1} { s}^{-1} \] Solving for \( \frac{d[SO_3]}{dt} \): \[ \frac{d[SO_3]}{dt} = 2 \times (2 \times 10^{-4}) \] \[ = 4 \times 10^{-4} { mol L}^{-1} { s}^{-1} \] Thus, the correct answer is (B).