Question

The reaction A + B → 2 C has the rate law rate = k[A]³. By what factor does the rate of reaction increase when [A] remains constant but is doubled?

Answer 1

So the rate law is given as \(r=k\left[A\right]\left[B\right]^{3}\)
It is saying that [A] is constant and is doubled. We can represent the new concentrations as [2B]. Now just substitute in the original rate law. \(r=k\left[A\right]\left[2B\right]^{3}\) You can simplify that to \(r=8k\left[A\right]\left[B\right]^{3}\) Then substitute r back in to get the new rate law as \(8r\)

Answer 2

the concentration of B is doubled*

it goes boldface every time I do [ B ]