Chemistry · Ch 6 — Chemical Kinetics

Order of the reaction

6.3.3

Order of the reaction

The overall order of a reaction is defined as the sum of the exponents in its experimentally determined rate law: for $\text{rate} = k[A]^x[B]^y$ , the overall order is $x+y$ . For example, $2\,H_2O_2(g) \rightarrow 2\,H_2O(l) + O_2(g)$ with rate $=k[H_2O_2]$ is first order overall, while $H_2(g)+I_2(g) \rightarrow 2\,HI(g)$ with rate $=k[H_2][I_2]$ is first order in each reactant and hence second order overall. Some key properties of reaction order, all illustrated in the textbook: (a) order is always an experimentally determined quantity, never assumed; (b) it can be an integer OR a fraction -- for $CH_3CHO(g) \rightarrow CH_4(g)+CO(g)$ th …

Misc Problem 6.4Order of 2 NO(g) + 2 H2(g) -> N2(g) + 2 H2O(g)

Worked out. For 2 NO(g) + 2 H2(g) -> N2(g) + 2 H2O(g), the experimental rate law is given as rate = k[NO]^2[H2]. Reading the exponents directly off this rate law: the power of [NO] is 2, so the reaction is second order in NO; the power of [H2] is 1, so it is …

Misc Problem 6.5Rate constant for A + B -> P (first order in A, second order in B)

Worked out. For A + B -> P, stated to be first order in A and second order in B, the rate law is rate = k[A][B]^2. Given rate = 3.6e-2 mol dm-3 s-1 when [A] = 0.2 mol dm-3 and [B] = 0.1 mol dm-3, solve for k: k = rate/([A][B]^2) = (3. …

Misc Problem 6.6Determining order in each reactant for A + B -> P from rate-change data

Worked out. Doubling [A] at constant [B] doubles the rate, so 2 = 2^x giving x = 1 (first order in A). Tripling [A] and doubling [B] together increases the rate 6-fold: 6 = 3^x x 2^y = 3 x 2^y, so 2^y = 2 and y …

You're reading a free preview — subscribe for the full notes →

← Previous · 6.3.2Writing the rate law Next · 6.4 →Molecularity of elementary reactions

Order of the reaction

6.3.3