Chemistry HW 4 - ekk86 hw04 Holcombe (53570) 1 This...

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

View Full Document Right Arrow Icon
ekk86 – hw04 – Holcombe – (53570) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When a reaction reaches equilibrium, 1. the forward and reverse reaction rates are equal. correct 2. the rate of the forward reaction and the rate of the reverse reaction are equal to zero. 3. all reaction stops. 4. the rate of the reverse reaction is zero. 5. the rate of the forward reaction is zero. Explanation: Chemical equilibrium is a dynamic equilib- rium in which both the forward and reverse reaction are continually going. There is a con- stant turn over of reactant to product and vice versa. By definition the two rates must equal one another in order for the concentrations of all species to remain constant with time. 002 10.0 points Write the equilibrium expression for the reac- tion 2 NO(g) + O 2 (g) y Y 2 NO 2 (g) 1. K = P 2 NO P O 2 P 2 NO 2 2. K = P NO 2 P 2 NO P O 2 3. K = 2 P NO 2 2 P NO P O 2 4. K = P 2 NO 2 P 2 NO P O 2 correct 5. K = P NO P NO 2 P O 2 Explanation: Since this is a gas phase reaction, K is the ratio of the partial equilibrium pressures of the products to the partial equilibrium pres- sures of the reactants. 003 10.0 points For the heterogeneous reaction 2 MnO 2 (s) y Y 2 MnO(s) + O 2 (g) , the equilibrium constant expression for K c is which of the following? 1. K c = [O 2 ] correct 2. K c = k [MnO 2 ] 3. K c = [MnO] [O 2 ] [MnO 2 ] 4. K c = 1 [O 2 ] 5. K c = [MnO 2 ] [MnO] [O 2 ] Explanation: For heterogeneous reactions, only gaseous reactants and species are taken into considera- tion when writing K c expressions. Therefore, K c = [O 2 ]. 004 10.0 points Given the following equilibria and equilibrium constants K 1 CO(g) + H 2 O(g) y Y CO 2 (g) + H 2 (g) K 2 CH 4 (g) + H 2 O(g) y Y CO(g) + 3 H 2 (g) K 3 CH 4 (g) + 2 H 2 O(g) y Y CO 2 (g) + 4 H 2 (g) The correct expression for K 3 in terms of K 1 and K 2 is 1. K 3 = K 1 + K 2 2. K 3 = K 1 - K 2 3. Cannot be determined from this informa- tion. 4. K 3 = K 1 K 2 correct 5. K 3 = K 1 K 2 Explanation:
Background image of page 1

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

View Full DocumentRight Arrow Icon
ekk86 – hw04 – Holcombe – (53570) 2 CO(g) + H 2 O(g) y Y CO 2 (g) + H 2 (g) K 1 = [CO 2 ][H 2 ] [CO][H 2 O] CH 4 (g) + H 2 O(g) y Y CO(g) + 3 H 2 (g) K 2 = [CO][H 2 ] 3 [CH 4 ][H 2 O] Adding (1) and (2), CO(g) + H 2 O(g) + CH 4 (g) + H 2 O(g) y Y CO 2 (g) + H 2 (g) + CO(g) + 3 H 2 (g) CH 4 (g) + 2 H 2 O(g) y Y CO 2 (g) + 4 H 2 (g) K 3 = [CO 2 ][H 2 ] 4 [CH 4 ][H 2 O] 2 K 1 K 2 = [CO 2 ][H 2 ] [CO][H 2 O] ± – [CO][H 2 ] 3 [CH 4 ][H 2 O] ± = [CO 2 ][H 2 ] 4 [CH 4 ][H 2 O] 2 = K 3 005 10.0 points The equilibrium constant for the reaction HNO 2 (aq) + H 2 O( ± ) NO - 2 (aq) + H 3 O + (aq) is 4 . 3 × 10 - 4 at 25 C. Will nitrous acid spon- taneously dissociate when [HNO 2 (aq)] = 0.15 M and [NO - 2 (aq)] = [H 3 O + (aq)] = 1 . 0 × 10 - 2 M? 1.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/01/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

Page1 / 8

Chemistry HW 4 - ekk86 hw04 Holcombe (53570) 1 This...

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