Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

give the relative rates of disappearance of reactants and formation for each of

ID: 485876 • Letter: G

Question

give the relative rates of disappearance of reactants and formation for each of the following reactions S T U D (D Cengage Learning/Carles D. Winters denotes challenging questions fully worked questions have answers in Appendix N and solutions in the student Solutions Manual 5. Experiment Time (s) Practicing skills 0.00 Reaction Rates 10.0 (See Section 14-1 Examples 14.1-14.2.) and 20.0 30.0 1. Give the relative rates of disappearance of reactants and 40.0 formation of products for each of the following reactions (a) Prep with (a) 2 O3(g) 3 O2 (g) of 2 HOF (g) 2 HF(g) O2(g) 40 2. Give the relative rates of disappearance of and formation of products for each of the following On (b) H reactions (a) 2 NO(g) Brz (g) NOBr(g) (b) N2 (g) 3 H2(g) 2 NH30g) 3. In the reaction 2 0,(g) 3 O2 (g), the rate of formation (c) of O2 is 1.5 x 10-3 mol/L s. at is the rate of decom- position of O3? 4. In the synthesis of ammonia, if -ALH2/At 4.5 x 10-4 molL min, what is ADNH3)/At? N2 (g) 3 H2 (g) 2 NH3(g) Unless otherwise noted, all art on this page is o Cengage Learning 2015.

Explanation / Answer

1)a) 2O3 ------> 3 O2

The relative rates of disppearance and formation are

- 1/2 d[O3] /dt = 1/3 d[O2]/dt

b) 2HOF ------> 2Hf + O2

-1/2 d[HOF]/dt = 1/2 d[HF]/dt = d[O2]/dt

Q2)

a) 2NO +Br2 ------> 2NOBr

-1/2 d[NO]/dt = -d[Br2]/dt = 1/2 d[NOBr]/dt

b) N2 + 3H2 ----> 2NH3

-1/2 d[N2]/dt = -1/3 d[H2]/dt = 1/2 d[NH3]/dt

3) For the given reaction

2O3 ------> 3 O2

The relative rates of disppearance and formation are

- 1/2 d[O3] /dt = 1/3 d[O2]/dt

hence

- d[O3] /dt = 2x1/3 d[O2]/dt

Given d[O2]/dt = 1.5 x10-3mol/L.s

Thus - d[O3]/dt = (2/3 ) x1.5 10-3mol/L.s

= 1.0 x 10-3mol/L.s

Q4)

N2 + 3H2 ----> 2NH3

-1/2 d[N2]/dt = -1/3 d[H2]/dt = 1/2 d[NH3]/dt

Thus

-1/3 d[H2]/dt = 1/2 d[NH3]/dt

or d[NH3]dt = 2/3 d[H2]dt

= (2/3) 4.5x10-4 mol/Lmin

= 3.0x 10-4 mol/L.min