In a reaction with a rate constant k, if the concentration of reactant A is doub
Practice Questions
Q1
In a reaction with a rate constant k, if the concentration of reactant A is doubled, how does the rate change if the reaction is second order with respect to A?
Rate remains the same
Rate doubles
Rate quadruples
Rate increases by a factor of eight
Questions & Step-by-Step Solutions
In a reaction with a rate constant k, if the concentration of reactant A is doubled, how does the rate change if the reaction is second order with respect to A?
Step 1: Understand that a second order reaction means the rate depends on the concentration of A squared.
Step 2: Write the rate equation for a second order reaction: Rate = k[A]^2.
Step 3: If the concentration of A is doubled, we replace [A] with 2[A].
Step 4: Substitute 2[A] into the rate equation: Rate = k(2[A])^2.
Step 5: Calculate (2[A])^2, which equals 4[A]^2.
Step 6: Now the new rate equation is Rate = k * 4[A]^2.
Step 7: Compare the new rate (4 times the original rate) to the original rate: The rate increases by a factor of 4.
Step 8: Conclude that if the concentration of A is doubled, the rate quadruples.
