Question:medium

For the reaction \( A + B \to C \), the rate law is found to be \( \text{rate} = k[A]^2[B] \). If the concentration of \( A \) is doubled and \( B \) is halved, by what factor does the rate change?

Show Hint

When concentrations change, apply exponent in rate law carefully to find new rate.

Updated On: Nov 26, 2025
  • \( 2 \)
  • \( \frac{1}{2} \)
  • \( 4 \)
  • \( 1 \)