Question

For a zero-order reaction, where \(k = 1.0\ mol\ L^{-1}\ min^{-1}\). If the initial concentration of A is \(2\ M\), then the time taken for completion of \(75\%\) of the reaction will be

• A. \(2.0\ min\)
• B. \(1.5\ min\)
• C. \(0.75\ min\)
• D. \(1.0\ min\)

Hint:

For a zero-order reaction, \[ [A]_t=[A]_0-kt \] Always calculate the remaining concentration first and then substitute into the integrated rate law.

Answer 1

The Correct Option is B

Concept: For a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. The integrated rate law for a zero-order reaction is: \[ [A]_t=[A]_0-kt \] where

• \([A]_0\) = initial concentration,

• \([A]_t\) = concentration after time \(t\),

• \(k\) = zero-order rate constant,

• \(t\) = time.

Step 1: Calculate the amount reacted. Initial concentration: \[ [A]_0=2.0\ M \] Given that \(75\%\) of the reaction is completed. Therefore, concentration consumed is \[ \frac{75}{100}\times2.0 \] \[ =1.5\ M \]

Step 2: Calculate the concentration remaining. Remaining concentration: \[ [A]_t=2.0-1.5 \] \[ =0.5\ M \]

Step 3: Apply the zero-order rate equation. Using \[ [A]_t=[A]_0-kt \] Substituting the given values: \[ 0.5=2.0-(1.0)t \] \[ t=2.0-0.5 \] \[ t=1.5\ min \]

Step 4: Verify the result. Since the rate constant is \(1.0\ mol\ L^{-1}\ min^{-1}\), consumption of \(1.5\ mol\ L^{-1}\) reactant should require exactly \(1.5\) minutes. Thus the answer is consistent. \[ \boxed{1.5\ min} \]