chem 1202 Group Final Exam Profs. Hogan, Kolniak, Stanley ANSWER KEY 2021

1202 Group Final Exam Profs. Hogan, Kolniak, Stanley
December 14, 2021 ANSWER KEY

Please answer all questions on your scantron answer sheet to hand in. You may keep these
question sheets. Each question has the same value.

1. Consider the following reaction:
A (aq) + B (aq) C (g)
At 300 K ΔHºrxn = – 400 kJ/mol and ΔSºrxn = 400 J/K-mol. Given this data select the
TRUE answer below.
G = H – TS; a negative
A. The reaction is spontaneous and Suniv > 0. Hrxn (exothermic) and
B. The reaction is endothermic and it’s becoming more ordered. positive
Srxn (increasing entropy
C. The reaction is endothermic and Suniv > 0 of reaction) will always
produce a negative
D. The reaction is nonspontaneous and Suniv < 0. Grxn, which is
spontaneous and
E. The reaction is becoming disordered and Suniv < 0 satisfies the second law
of thermodynamics with


2. What is the melting point, C, for A, given ΔHºfusion = 15 kJ/mol and ΔSºfusion = 150 J/K-
mol.
A (s) A (l)

a) 373°C b) 298°C c) 100°C d) –100°C e) –173°C

Tfusion(melting )
Hfusion  15 kJ/mol  100 K; T (C) = 100  273 =  173 C Remember to
 convert S to
Sfusion 0.15 kJ/K mol
match units on




3. From the thermodynamic data given below, calculate Srxn for the following reaction:
Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)

a) 210 J/K•mol b) 90 J/mol·K) c) –90 J/K•mol d) –210 J/K•mol
J/K•mol
Sf( e) –250 J/K•mol
Mg(OH)2 (s) 60 Srxn = Sprod  Sreact
Srxn = [ (90 J/mol•K) + (2)(70)]  [ (60 J/mol•K) + (2)(190 J/mol•K)]
HCl (aq) 190 Srxn = 230 J/mol•K  440 J/mol•K = 210 J/mol•K
MgCl2 (aq) 90
H2O (l) 70

, 4. Consider the following generic reaction: A  B, for which Keq = 0.08 at 100°C. Calculate
G°rxn for this reaction at 100°C? R = 8.314 J/K-mol
a) 7.8 kJ/mol b) 2.1 kJ/K•mol c) –2.1 kJ/K•mol d) –7.8 kJ/K•mol e) –10.8 kJ/K•mol

G = RTln(Keq) = (8.314 J/K•mol)(373K)ln(0.08) = (3101 J/mol)(2.52) = 7814 J/mol = +7.8 kJ/mol


5. A certain reactant A follows first order kinetics with a half-life, t1/2, = 0.25 s. How long will
it take the initial concentration of A to be reduced to 12.5%?

a) 0.25 s b) 0.50 s c) 0.75 s d) 1.0 s e) 1.25 s
One half-life (0.25 s) takes A from 100% to 50%, the second half-life (0.5 s) cuts the amount of A to
25%. The third half-life (0.75 s) cuts 25% in half to 12.5%. So three half-lives = 0.75 s

6. With regard to kinetics, which of the following statements is (are) True?
I. Increasing the temperature decreases the rate constant k.
II. The addition of a catalyst increases the rate constant k.
III. Decreasing the activation energy barrier, Ea, increases the rate constant k.

A. I only (I) is FALSE as increasing the temperature increases
B. II only the rate of reaction and the rate constant.
(II) is TRUE as adding a catalyst increases the rate of
C. III only reaction and the rate constant.
(III) is TRUE as reducing the activation barrier (what
D.II and III only a catalyst does) increases the rate of reaction
and the rate constant.
E. All are True.
7. Consider the following reaction: A (aq) + B (aq) + C (aq) D (aq) + E (l)
Determine the rate law for this reaction using the data in the following table:

Exp # [A] (M) [B] (M) [C] (M) Rate
(M•sec1)
1 0. 0. 0. 0.03
2 2 1
2 0. 0. 0. 0.12
4 2 1
3 0. 0. 0. 0.12
4 4 1
4 0. 0. 0. 0.09
2 2 3

a) rate = k[A][B][C] [A]: exp #1 to #2, doubling [A] while leaving [B] and [C] the same leads to

b) rate = k[A][B]  [A2]  x   Rate2 ;  [0.4]  x   0.12 ; 2 x   4  ; x = 2
 [A ]  Rate   [0.2]  0.03 
1   1   
: exp #3 to #2, doubling [B] while leaving [A] and [C] the same leads to NO
: exp #1 to #4, tripling [C] while leaving [A] and [B] the same leads to a tr
The rate law based on this data is: rate = k[A]2[C]