CHhmwk10

# CHhmwk10 - moore(jwm2685 Homework 10 vanden bout(51640 This...

This preview shows pages 1–3. Sign up to view the full content.

moore (jwm2685) – Homework 10 – vanden bout – (51640) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Consider the mechanism NO 2 + ± 2 NO 2 ± + ± k 1 , slow ± + NO 2 NO 2 ± k 2 , Fast What is the rate law? 1. rate = k 2 [NO 2 ] 2 2. rate = k 2 [NO 2 ] [±] 3. rate = k 1 [NO 2 ] [± 2 ] correct 4. rate = k 1 [NO 2 ±] [±] 5. rate = k 1 k 2 [NO 2 ] 2 Explanation: 002 10.0 points Consider the mechanism ClO (aq) + H 2 O( ) HOCl(aq) + OH (aq) K , Fast I (aq) + HOCl(aq) HOI(aq) + Cl (aq) k 1 , slow HOI(aq) + OH (aq) OI (aq) + H 2 O( ) k 2 , Fast What is the rate law? 1. rate = k 1 K [ClO ] [I ] [OH ] 2. rate = k 1 k 2 K [ClO ] [I ] 3. rate = k 1 K [ClO ] [I ] [OH ] 1 correct 4. rate = k 1 K [ClO ] [I ] 5. rate = k 1 [I ] [HOCl] Explanation: 003 10.0 points The reaction 2 NO 2 (g) 2 NO(g) + O 2 (g) is postulated to occur via the mechanism NO 2 (g) + NO 2 (g) NO(g) + NO 3 (g) slow NO 3 (g) NO(g) + O 2 (g) Fast. What is an intermediate in this reaction? 1. NO 2 (g) 2. NO(g) 3. ON - NO 3 (g) 4. O 2 (g) 5. NO 3 (g) correct Explanation: 004 10.0 points Determine the overall balanced equation For a reaction having the Following proposed mech- anism Step 1: B 2 + B 2 -→ E 3 + D slow Step 2: E 3 + A -→ B 2 + C 2 Fast and write an acceptable rate law. 1. A + B 2 -→ C 2 + D; R = k [B 2 ] 2 correct 2. E 3 + A -→ B 2 + C 2 ; R = k [E 3 ] [A] 3. A + B 2 -→ C 2 + D; R = k [A][B 2 ] 4. B 2 + B 2 -→ E 3 + D; R = k [B 2 ] 2 Explanation: Step 1: B 2 + B 2 -→ E 3 + D slow Step 2: E 3 + A -→ B 2 + C 2 Fast balanced equation, rate law = ? A + B 2 -→ C 2 + D (From the 2 molecules oF B 2 in the rate- determining step) 005 10.0 points Consider the reaction mechanism below: Step Reaction 1 2H 2 S + 3O 2 -→ 2SO 2 + 2H 2 O 2 2H 2 S + SO 2 -→ 3S + 2H 2 O 3 2S -→ S 2 overall 2H 2 S + O 2 -→ S 2 + 2H 2 O

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
moore (jwm2685) – Homework 10 – vanden bout – (51640) 2 Assume step 3 is the rate-determining step. Which rate law below could describe this re- action? 1. rate = k · [H 2 S] · [O 2 ] · [H 2 O] 2 2. rate = k · [H 2 S] 2 · [O 2 ] · [H 2 O] 2 cor- rect 3. rate = k · [H 2 S] · [O 2 ] · [H 2 O] 1 4. rate = k · [H 2 S] 2 · [O 2 ] · [H 2 O] 2 5. rate = k · [H 2 S] 2 · [O 2 ] Explanation: Canceling the intermediates for this reac- tion is tricky and can be approached in sev- eral ways. The easiest starting point is to ensure that the coeFcients in the steps match those in the balanced equation. Starting from S 2 (since it appears only once in the mecha- nism), it is clear that we needn’t alter step 3. Canceling the intermediate S therefore requires multiplying step 2 by 2/3. Conse- quently, canceling the intermediate SO 2 re- quires multiplying step 1 by 1/3. 006
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 05/02/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

### Page1 / 9

CHhmwk10 - moore(jwm2685 Homework 10 vanden bout(51640 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online