Question

A → P is a zero-order reaction. At 298 K, the rate constant of the reaction is \( 1 \times 10^{-3} \) mol L\(^{-1}\) s\(^{-1}\). Initial concentration of 'A' is 0.1 mol L\(^{-1}\). What is the concentration of 'A' after 10 sec?

• A. 0.09 mol L\(^{-1}\)
• B. 0.099 mol L\(^{-1}\)
• C. 0.087 mol L\(^{-1}\)
• D. 0.011 mol L\(^{-1}\)

Hint:

For a zero-order reaction, the concentration decreases linearly with time: \[ [A] = [A]_0 - k t \]

Answer 1

The Correct Option is A

Step 1: Zero-Order Reaction Formula For a zero-order reaction, the concentration at time \( t \) is given by: \[ [A] = [A]_0 - k t \] Step 2: Substituting the Given Values \[ [A] = 0.1 - (1 \times 10^{-3} \times 10) \] \[ [A] = 0.1 - 0.01 = 0.09 \text{ mol L}^{-1} \] Thus, the correct answer is 0.09 mol L\(^{-1}\).

Answer 2

Given:
- Reaction: A → P is zero-order
- Rate constant, \( k = 1 \times 10^{-3} \) mol L\(^{-1}\) s\(^{-1}\)
- Initial concentration of A, \( [A]_0 = 0.1 \) mol L\(^{-1}\)
- Time, \( t = 10 \) s

For a zero-order reaction, the concentration of reactant A at time t is given by the equation:
\[ [A]_t = [A]_0 - kt \]

Substitute the values:
\[ [A]_{10} = 0.1 - (1 \times 10^{-3}) \times 10 = 0.1 - 0.01 = 0.09 \text{ mol L}^{-1} \]

Therefore, the concentration of A after 10 seconds is:
\[ \boxed{0.09 \text{ mol L}^{-1}} \]