Question

At \(1130\,K\), the decomposition of ammonia on Pt catalyst follows zero order kinetics. The rate of this reaction at \(t=10\) min is \(x\ mol\,L^{-1}\,min^{-1}\). What will be its rate (in \(mol\,L^{-1}\,min^{-1}\)) at \(t=20\) min, at the same temperature?

• A. \[ \frac{x}{2} \]
• B. \[ x \]
• C. \[ 2x \]
• D. \[ \sqrt{x} \]

Hint:

For a zero-order reaction: \[ \text{Rate}=k \] \[ [A]_t=[A]_0-kt \] The rate remains constant throughout the reaction and is independent of reactant concentration.

Answer 1

The Correct Option is B

Concept: For a zero-order reaction, \[ \text{Rate}=k \] where \(k\) is the zero-order rate constant. Thus, the rate is independent of the concentration of reactants and remains constant with time.

Step 1: Write the rate law for a zero-order reaction. \[ \text{Rate}=k[A]^0 \] \[ \text{Rate}=k. \]

Step 2: Compare the rates at different times. Since the rate does not depend on concentration, \[ \text{Rate at } t=10\ \text{min} = \text{Rate at } t=20\ \text{min}. \] Given, \[ \text{Rate at } t=10\ \text{min} = x\ mol\,L^{-1}\,min^{-1}. \] Therefore, \[ \text{Rate at } t=20\ \text{min} = x\ mol\,L^{-1}\,min^{-1}. \]

Final Answer: \[ \boxed{x\ mol\,L^{-1}\,min^{-1}} \] \[ \boxed{\text{Answer = (B)}} \]