{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}


Exam_2_2008_Solutions - CHEM 142 Solutions Exam 2 April 4...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Page 1 of 6 CHEM 142 Solutions Exam # 2 April 4, 2008 1. ( 10 pts ) Consider the following reaction that is initially at equilibrium. Predict which way the reaction will proceed for each of the disturbances to equilibrium listed below. Explain each briefly : 2NO(g) + Cl 2 (g) 2 NOCl(g) Δ H = - 85 kJ/mole a.) The partial pressure of NOCl(g) is doubled ANS: Doubling the NOCl partial pressure is equivalent to adding more NOCl. This will drive the reaction in the reverse direction toward reactants. b.) 1 atm of Ar is added to the equilibrium mixture at fixed Volume ANS: Adding a gas which is not involved in the reaction will not shift the equilibrium at all so no change will occur. c.) The temperature is increased from 298 K to 350 K ANS: The Δ H = - for the reaction, this makes it exothermic. Increasing the temperature favors endothermic reactions, which is the reverse. Therefore the reaction will go in reverse. 2. ( 10 pts ) Determine the value for the equilibrium constant, K, for the following reaction given that for HF K a = 3.5 x 10 -4 and for HCN K a = 4.9 x 10 -10 . Please show your work. ( Hint: although this is an acid base reaction use the approaches developed for general equilibrium in chapter 14) HCN(aq) + F - (aq) F HF(aq) + CN - (aq) K = ?? ANS: This reaction can be obtained by combining together two reactions involving the acid ionization of HF and HCN. The component reactions are: HCN(aq) F H + (aq) + CN - (aq) K a1 = 4.9 x 10 -10 (1) HF(aq) + F H + (aq) + F - (aq) K a2 = 3.5 x 10 -4 (2) In order to get the original reaction we need to reverse reaction (2). When this reaction is reversed the equilibrium constant of the reversed reaction is the reciprocal of the original equilibrium constant. HCN(aq) F H + (aq) + CN - (aq) K a1 = 4.9 x 10 -10 (1) H + (aq) + F - (aq) F HF(aq) K 3 = 1/ K a2 = (1/3.5 x 10 -4 ) = 2.87 x10 3 (3) The target reaction is the sum of these two. When we sum two reactions the equilibrium constant of the resulting reaction is the product of the first two.
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}