Question

The rate of a reaction \( A \rightarrow B \) is \( 0.2 \, \text{mol} \, \text{m}^{-3} \, \text{s}^{-1} \) at a particular concentration \( C_{A1} \). The rate constant of the reaction at a given temperature is \( 0.1 \, \text{m}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \). If the reactant concentration is increased to \( 10 \, C_{A1} \) at the same temperature, the reaction rate, in \( \text{mol} \, \text{m}^{-3} \, \text{s}^{-1} \), is:

• A. 20
• B. 10
• C. 100
• D. 50

Hint:

For reaction rate problems, determine the initial concentration from the rate law and apply the proportional relationship between the rate and reactant concentration for changes in \( C_A \).

Answer 1

The Correct Option is A

Given: \[ A \xrightarrow{k} B \] \[ - r_{A_1} = 0.2 \, \text{mol/m}^3 \cdot \text{sec at } C_{A_1}, \] \[ k = 0.1 \, \text{m}^3/\text{mol} \cdot \text{sec}, \quad \text{find } r_{A_2} \text{ at } 10C_{A_1}. \] Step 1: Determining the reaction order. From the rate constant unit, we can deduce the reaction order: \[ k \, \text{unit} = \left( \frac{\text{mol}}{\text{lit}} \right)^{1-n} \text{sec}^{-1}. \] For \( k = 0.1 \, \text{m}^3/\text{mol} \cdot \text{sec} \): \[ 1 - n = -1, \quad n = 2. \] Step 2: Rate equation for the reaction. \[ - r_{A_1} = 0.1 C_{A_1}^2. \] Step 3: Calculating \( r_{A_2} \). \[ \frac{-r_{A_2}}{-r_{A_1}} = \frac{0.1 C_{A_2}^2}{0.1 C_{A_1}^2} \quad \text{(since \( k \) remains constant as the temperature remains constant)}. \] \[ C_{A_2} = 10C_{A_1}. \] \[ \frac{-r_{A_2}}{0.2} = \left(\frac{10C_{A_1}}{C_{A_1}}\right)^2. \] \[ - r_{A_2} = 20 \, \text{mol/m}^3 \cdot \text{sec}. \] Final Answer: \[ - r_{A_2} = 20 \, \text{mol/m}^3 \cdot \text{sec}. \]