The reaction A + 2B + C --> Products has a rare law of Rate = k[A]^2[B]. By what factor would the reaction rate change if [B] is tripled?

The reaction  A+B YIELDS C+D rate=k[A][B]^2 has an initial rate of 0.0970 M/s. What will the initial rate be if [A] is halved and [B] is tripled?

Consider the reaction and its rate law. 3A+3B⟶productsrate=?[A][B]2 What is the order with respect to A?

4. The rate law for the reaction H₂ (g) + I₂ (g) → 2 HI (g) was determined to be rate = k [H₂] [1₂], which led to a simple single-step bimolecular mechanism where an H₂ molecule simply collided with an I₂ molecule. In the 1960s, though, different teams found spectroscopic evidence for this mechanism: (1) (2) (3) I₂ (g) →21 (g) H₂(g) + I (g) → H₂I (g) H₂I (g) + I (g) → 2 HI (g) (fast) (fast) (slow) Show that this mechanism is consistent with the rate law: rate = k [H₂] [1₂]. To do this, assign individual rate constants for ALL forward and reverse reactions in the mechanism (I'm guessing k₁, k-1, k2, k-2 and k3). Then show that the overall rate law from combining individual rate laws for the mechanism steps yields rate = (some combination of k's) [H₂] [1₂], and thus k= (some combination of k's). Remember that there should be no reaction intermediates in the rate law, only reactants. \