FINAL-solutions

# FINAL-solutions - Version 174 – FINAL – sutcliffe...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 174 – FINAL – sutcliffe – (51045) 1 This print-out should have 38 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Calculate the constant A in the Arrhenius equation, given that the second order reac- tion rate constant for a reaction is 1.00 × 10 3 L/mol · s at 500 ◦ C and the activation energy E a is 19.87 kcal/mole. 1. 1 . 00 × 10 7 2. 1 . 00 × 10 8 3. 4 . 55 × 10 3 4. 4 . 15 × 10 8 correct 5. 2 . 4 × 10 6 Explanation: k = 1 . × 10 3 L · s mol E a = 19870 cal mol T = 500 ◦ C + 273 = 773 K k = Ae − E a /RT A = ke E a /RT = ( 1 × 10 3 ) exp bracketleftbigg 19870 (1 . 987)(773) bracketrightbigg = 4 . 15 × 10 8 002 10.0 points NOTE: If this helps, the K a for HClO 3 is ENORMOUS!! All components are present in 0.10 M con- centrations. I) HCN and NaCN II) NH 3 and NH 4 Cl III) HNO 3 and NH 4 NO 3 IV) HClO 3 and NaClO 3 Which will give buffer solutions? 1. I and III only 2. I and II only correct 3. II, III and IV only 4. I, III and IV only 5. III and IV only Explanation: Buffers are formed in one of two ways, by combining a weak acid and its conjugate base or by combining a weak base and its conjugate acid. HNO 3 and HClO 3 are both strong acids and cannot be used to make effective buffer solutions. HCN is a weak acid and NaCN is the salt of its conjugate base, CN − . NH 3 is a weak base and NH 4 Cl is the salt of its conjugate acid, NH + 4 . Therefore 1 and 2 can be used to make effective buffer solutions. 003 10.0 points HINT: If you’re having trouble figuring out the order for one of these then remember: your choices for each reagent are first, second and third. Try trial and error! Three separate experiments were per- formed on the rate of the reaction 3 A 2 + 2 B → 2 A 3 B. The measured initial concentrations of A 2 (in moles per liter) are shown below along with the measured initial rates of formation of A 3 B (moles per liter per second). Initial Initial Initial Trial [A 2 ] [B] rate M M M/s 1 1 . 2 2 . 4 8 . × 10 − 8 2 1 . 2 1 . 2 4 . × 10 − 8 3 1 . 8 2 . 4 1 . 8 × 10 − 7 What is the order of the reaction? 1. first order in [A 2 ] and first order in [B] 2. second order in [A 2 ] and first order in [B] correct 3. third order in [A 2 ] and second order in [B] Version 174 – FINAL – sutcliffe – (51045) 2 4. first order in [A 2 ] and second order in [B] 5. None of these is correct. Explanation: Rate = k [A 2 ] x [B] y Rate 3 Rate 1 = k [A 2 ] x 3 [B] y 3 k [ A 2 ] x 1 [ B ] y 1 1 . 8 × 10 − 7 8 . × 10 − 8 = parenleftbigg 1 . 8 1 . 2 parenrightbigg x parenleftbigg 2 . 4 2 . 4 parenrightbigg y 2 . 25 = 1 . 5 x ln 2 . 25 = x ln 1 . 5 x = ln 2 . 25 ln 1 . 5 = 2 Rate 1 Rate 2 = k [A 2 ] 2 1 [B] y 1 k [ A 2 ] 2 2 [ B ] y 2 8 . × 10 − 8 4 . × 10 − 8 = parenleftbigg 1 . 2 1 . 2 parenrightbigg 2 parenleftbigg 2 . 4 1 . 2 parenrightbigg y 2 1 = 2 y y = 1 004 10.0 points Consider a reaction for which (at 300 K) Δ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 12

FINAL-solutions - Version 174 – FINAL – sutcliffe...

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

View Full Document
Ask a homework question - tutors are online