HW 3 - – Homework#3 – Holcombe – 1 This print-out...

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: – Homework #3 – Holcombe – 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the reaction CaCO 3 (s) ⇀ ↽ CaO(s) + CO 2 (g) , at equilibrium at a certain temperature. Now a small amount of Ca 14 CO 3 (s) ( 14 C la- beled calcium carbonate) is added. What species will be present after equilibrium is re-established? 1. CaCO 3 (s), CaO(s), 14 CO 2 (g), CO 2 (g) 2. Ca 14 CO 3 (s), CaO(s), 14 CO 2 (g), CO 2 (g) 3. CaCO 3 (s), Ca 14 CO 3 (s), CaO(s), 14 CO 2 (g), CO 2 (g) correct 4. CaO(s), 14 CO 2 (g), CO 2 (g) 5. CaCO 3 (s), Ca 14 CO 3 (s), CaO(s), CO 2 (g) Explanation: The addition of extra reactant Ca 14 CO 3 will favor the production of more products (CaO(s) and 14 CO 2 (g)). However, since this is a dynamic equilibrium, both forward and backward reactions occur. Therefore, an equilibrium will be established where 14 C is present in both products and reactants con- taining C. 002 10.0 points In which of the following cases will the LEAST time be required to arrive at equilibrium? K below refers to the equilibrium constant. 1. K is about 1. 2. K is a very large number. 3. Cannot tell without knowing the value of K . 4. Cannot tell since the time to arrive at equilibrium does not depend on K . correct 5. K is a very small number. Explanation: The equilibrium constant K reflects the relative amounts of products and reactants present when the reaction is complete, and gives no information on the time needed. The rate constant k indicates how fast the reaction occurs. 003 10.0 points Using the law of mass action, write the equilibrium expression for the following reac- tion: 2 Cu 2+ (aq) + 4 I − (aq) ←→ 2 CuI(s) + I 2 (aq) 1. K = [Cu 2+ ] 2 [I − ] 4 [I 2 ][CuI] 2 2. K = [I 2 ] [Cu 2+ ] 2 [I − ] 4 correct 3. K = [Cu 2+ ] 2 [I − ] 4 [I 2 ] 4. K = [I 2 ][CuI] 2 [Cu 2+ ] 2 [I − ] 4 Explanation: The equilibrium expression K is the ratio of products to reactants raised to their respec- tive stoichiometric coefficients. Solid species are not included in the expression. 004 10.0 points Write the reaction quotient for MgSO 4 · 7 H 2 O(s) → MgSO 4 (s) + 7 H 2 O(g) 1. Q = P MgSO 4 P 7 H 2 O P MgSO 4 · 7H 2 O 2. Q = 1 P 7 H 2 O 3. Q = P 7 H 2 O P MgSO 4 · 7H 2 O – Homework #3 – Holcombe – ( 2 4. Q = P MgSO 4 · 7 P H 2 O P MgSO 4 · 7H 2 O 5. Q = P 7 H 2 O correct Explanation: Q = activities of products activities of reactants , where “activities” are for an ideal gas: a J = [ P J ] for a solute: a J = [J] for a pure solid or liquid: a J = 1 005 10.0 points Consider the following reactions at 25 ◦ C: reaction K c 2 NO(g) ⇀ ↽ N 2 (g) + O 2 (g) 1 × 10 30 2 H 2 O(g) ⇀ ↽ 2 H 2 (g) + O 2 (g) 5 × 10 − 82 2 CO(g) + O 2 (g) ⇀ ↽ 2 CO 2 (g) 3 × 10 91 Which compound is most likely to dissociate and give O 2 (g) at 25 ◦ C?...
View Full Document

{[ snackBarMessage ]}

Page1 / 7

HW 3 - – Homework#3 – Holcombe – 1 This print-out...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online