Final answer:
Doubling the concentration of reactants in a second-order reaction will quadruple the reaction rate, resulting in a new rate expression of 4K, where K is the initial rate constant. However, for a first-order reaction, doubling the reactant concentration simply doubles the reaction rate, leading to a new rate expression of 2K.
Step-by-step explanation:
When we discuss doubling a reaction and its relation to the rate constant (K), we are referring to the effect of changing reactant concentrations on the rate of reaction. If the reaction under consideration follows a second-order kinetics, such as 2HI → H2 + I2, then doubling the concentration of HI will increase the reaction rate by four times because the rate expression is rate = k[HI]². As a result, the new rate is calculated by k(2[HI])² = 4k[HI]², thus the rate constant in terms of observable rate would appear to be quadrupled, making the answer 4K.
For a first-order reaction, like the decomposition of cyclopropane, doubling the initial concentration of the reactant will double the reaction rate, since the rate expression is rate = k[cyclopropane]. This is consistent with the rate law for a first-order reaction. Therefore, if the initial concentration is doubled, the reaction rate would just become 2k[cyclopropane], doubling the observable rate.
In both cases, the rate law dictates how the rate of the reaction will change with respect to changes in reactant concentration, but the actual rate constant (k) remains unchanged. It is the observed rates that are impacted by the concentration changes, not the intrinsic rate constant itself.