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

Step 1: Write the zero-order rate law.
For a zero-order reaction \[ [A]_t = [A]_0 - kt \] where the rate does not depend on concentration.
Step 2: List the data.
\([A]_0 = 2.0\) M, \(k = 1.0\,mol\,L^{-1}\,min^{-1}\), and 75 percent of A reacts.
Step 3: Find the amount consumed.
\(0.75 \times 2.0 = 1.5\) M of A is used up.
Step 4: Find the concentration left.
\([A]_t = 2.0 - 1.5 = 0.5\) M.
Step 5: Substitute into the rate law.
\[ 0.5 = 2.0 - (1.0)\,t \implies t = 2.0 - 0.5 = 1.5\,min \]
Step 6: Sanity check.
Since \(k = 1.0\) means 1 mol/L is consumed each minute, removing 1.5 mol/L indeed needs 1.5 minutes.
\[ \boxed{1.5\,min} \]