The conversion of molecules to follows second order kinetics. If concentration of is increased to three times how will it affect the rate of formation of ?
For a second order reaction, the rate law is given by:
Rate = k[X]²
When the concentration of X is increased to three times, the new concentration becomes 3[X]. Substituting this in the rate equation:
New Rate = k[3X]² = k × 9[X]² = 9 × k[X]²
Therefore, the rate of formation of Y will increase by 9 times (or becomes nine times the original rate).
Explanation
This question tests understanding of how reaction rate depends on concentration for different orders. For a second order reaction, the rate is proportional to the square of the concentration of the reactant. When concentration is tripled (3 times), the rate increases by (3)² = 9 times. The textbook context shows similar examples where rate depends on square of concentration (like NO in the example given).
Solution Steps
Step 1: Write the rate law for second order reaction: Rate = k[X]²
Step 2: Substitute new concentration (3[X]) into the rate equation: New Rate = k[3X]²
Step 3: Simplify: New Rate = k × 9[X]² = 9 × k[X]² = 9 × original rate
Step 4: Conclude that rate increases by 9 times.