{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

from_F04_FED_key

from_F04_FED_key - Chemistry 111 KEY Page 1 out of 8 Final...

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

Chemistry 111 KEY Page 1 out of 8 Final Exam D / Fall 2004 1. Consider the following reaction… 2 A (g) + 3 B (g) 4 C (g) + D (g) Which of the following is the mathematical definition of Kc for this reaction? a) [A] 2 [C] 4 [D] b) [A] 2 [B] 3 [C] 4 [D] c) [C] 4 [D] [A] 2 d) [C] 4 [D] [A] 2 [B] 3 e) [A] [B] [C] [D] 2. Consider the reaction below with a Kc = 9.00. In a given 1-L flask, we place 0.150 mol of A, 0.150 mol of B, and 0.600 mol of C. What will be the concentration of C at equilibrium? A (g) + B (g) 2 C (g) a) 0.63 M b) 0.57 M c) 0.51 M d) 0.54 M e) None of these. First set up the initial conditions and calculate Q in order to decide the direction of the shift. A + B 2 C init: 0.15 M 0.15 M 0.60 M Calculation of Q is given by: Q= [A][B] [C] 2 = (0.15) 2 (0.60) 2 =1 6 Since Q > Kc, then the equilibrium will shift to the left until Q = 9.00. Therefore, A + B 2 C init: 0.15 M 0.15 M 0.60 M @ eq: 0.15 + x 0.15 + x 0.60 - 2x Substituting into the mathematical expression for Kc, we get: Kc = [A][B] [C] 2 = (0.15 + x) 2 (0.60 - 2x) 2 =9 . 0 0 Taking square roots on both sides, we get: 0.15 + x 0.60 - 2x =3 x = 0.03 M However, the answer must be 0.60 – 2(0.03) = 0.54 M .

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

View Full Document
Chemistry 111 KEY Page 2 out of 8 Final Exam D / Fall 2004 3. Consider the following reaction for which Kp = 2.50 × 10 -5 . In a given flask we place gases A, B, and C at initial pressures of 0.100, 0.200, and 0.300 atm, respectively. What will be the pressure of gas D at equilibrium? 2 A (g) + 3 B (g) 4 C (g) + 3 D (g) a) 1.59 torr b) 4.77 torr c) 9.01 torr d) 14.4 torr e) None of these. Once more, we'll start by examining the initial conditions and calculating Q to decide the direction of the shift. init: 0.100 atm 0.200 atm 0.300 atm 0 2 A + 3 B 4 C + 3 D Since the initial pressure of D is zero, then Q must be equal to zero. This means that we are very close to Kc from the beginning. So, we will shift to the right. How much? Just a tiny bit. This will make x negligible… init: 0.100 atm 0.200 atm 0.300 atm 0 2 A + 3 B 4 C + 3 D @ eq: 0.100 - 2x 0.200 - 3x 0.300 + 4x 3x Substitution into Kc yields: 43 4 3 C 23 2 3 [C] [D] (0.3) (3x) K= = = [A] [B] (0.1) (0.2) 2.50 x 10 -3 3x = [D] = 6.28 × 10 -3 atm. That is a small number so we will convert it to torr (also because those are the units given for the answer). -3 760torr 6.28x10 atm x 1atm = 4.77 torr . 31. Consider the reaction below with a Kc of 2.25 × 10 +12 . In a given 1-L flask, we place 2.00 mol of A, 8.00 mol of B, 2.00 mol of C and no D. What will be the concentration of gas A at equilibrium?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 8

from_F04_FED_key - Chemistry 111 KEY Page 1 out of 8 Final...

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

View Full Document
Ask a homework question - tutors are online