Chemistry 2000 Section B Midterm Test
Solutions Set
Question 1 (10 Marks, 8 marks for part a and 2 for part b)
a) Draw a sketch of the phase diagram of water, indicate the solid-liquid, solidvapor, and liquid-vapor equilibrium lines, the triple point and the critical point.
Also indicate the regions that represent the solid, liquid and vapor phases.
b
c
P
Critical
point
liquid
solid
vapor
a
Triple point
T
The equilibrium lines are as follows a- solid-vapor, b- sold-liquid and c-liquidvapor. The triple point is at 273.16 K and 0.00603 atm and critical point 647.6
K and 218 13 atm
b) With reference to the above diagram give an explanation of why water melts
when pressure is increase while for carbon dioxide this does not occur.
The liquid-solid equilibrium line in water has a negative slope and the same line in
carbon dioxide has a positive slope.
Question 2 (2 Marks Each)
True or False:
a) Hydrogen bonding in a solid decreases its melting point. False
b) The Boiling point of NH3 is greater than CH4. True
c) Hydrogen bonding interactions dominate in CH3F. False
d) The boiling point of Xe is higher than Ar. True
e) The process going from liquid to solid is exothermic. True
f) N2 is more soluble to O2 in H2O. False
g) NH3 forms a molecular solid. True
h) HF has a lower boiling point than HCl. False
i) Br2 is less polarizable than Cl2. False
j) Ionic compounds dissolve in water due to their strong dipole-dipole interactions.
False
Question 3 (20 Marks, 10 marks for part a and 10 marks for part b)
A solution is prepared with 250.00 g of a white solid in 200.00 g of water. The boiling
point of pure water is 100.000 0C, and the melting point of pure water is 0.000 0C.
a) Given that the boiling point is 100.735 0C, determine the freezing point of the solution.
∆TBP = TBP –TBppure = 100.735 0C - 100.000 0C = 0.735 0C
Msolute = ∆TBP/KBP = 0.735 0C/0.5121 0C = 1.435 mol kg-1
∆TFP = msolute*KFP = (1.435 mol kg-1)(-1.860 0C kg mol-1) = -2.670 0C
TFP = TFPpure + ∆TFP = 0.000 0C + -2.670 0C = -2.670 0C
b) Compute the osmotic pressure, in atm, of a solution of 136 g of this white solid in 124
g of water at 25.00 0C. Assume the density of this solution is 1.230 g/mL.
MMsolute = (msolute/M)/msolvent = (250.00 g/ 1.435 mol kg-1)/(0.20000 kg) = 871.1 g mol-1
nsolute = ms/MMsolute = 136 g/871.1 g mol-1 = 0.0156 mol
c = n/(m/d) = 0.156 mol/ ((136 g +124 g)/(1.250 g mL-1/1000 mL L-1)) = 0.751 mol L-1
Π = cRT = (0.751 mol L-1)(0.05206 L atm mol-1 K-1)(298.15 K) = 18.4 atm
Note: The subscript digits in the molar mass and the molarity are kept for the purpose of
avoiding excessive rounding errors in the calculation.
Question 4 (25 marks, part a) 3 marks, part b) 10 marks, part c) 5 marks, part d) 5 marks
and part e) 2 marks)
The inter-conversion of cis-2-butene to trans-2-butene is a first order process with
activation energy of 262 kJ mol-1.
a) Write the rate equation and the corresponding intergrated rate law for this reaction.
Rate = k[cis-2-butene]
(Rate equation)
[cis-2-butene] = [cis-2butene]0(exp(-kt))
(Integrated Rate Law)
or ln([cis-2-butene]/ [cis-2-butene]0) = -kt
b) The rate of the reaction at 298 K, when [cis-2-butene] = 0.25 M, is 1.3*10-15 M s-1.
Determine the rate of this reaction when the temperature is raised to 500 K when [cis-2butene] = 0.35 M.
Rate298 = k1*[cis-2-butene] = 1.3*10-15 M/s / 0.25 M = 5.2*10-15 s-1
ln k2 = ln k1 + Ea/R(1/T1-1/T2)
= ln(5.2*10-15) + 2.62*105 J/ 8.3145 J K-1 mol1 (1/298 K – 1/500 K)
= -32.98 + 31511(0.001356) = 9.829
Î k2 = 1.86*104 s-1
Rate500 = (1.86*104 s-1)*(0.35 M) = 6.51*103 M s-1
c) Determine the half-life of this reaction at 500 K.
t1/2 = 0.6931/k2 = 0.6931/1.86*104 s-1 = 3.73*10-5 s
d) Determine the concentration of cis-2-butene after 10 microseconds at 500 K when the
initial concentration of cis-2-butene is 1.5 M.
ln[cis-2-butene] = -k2t + ln[cis-2-butene]0
= -(1.86*104 s-1)(1.0*10-5) + ln(1.5 M)
=-0.1858 + 0.4054 = 0.2196 = ln [cis-2-butene]
Î [cis-2-butene] =1.25 M
e) When I2 is added the reaction speeds up greatly. Give an explanation for this
phenomenon.
Iodine is a catalyst for the overall reaction. It serves to lower the activation energy,
which in turn speeds up the reaction rate.
Question 5 (20 marks, part a) 4 marks, part b) 3 marks, part c) 8 marks and part d) 5
marks)
The Kc for the decomposition of ammonium hydrogen sulfide is 1.8*10-4 M2 at 250C, in a
1.00 L container.
NH4HS(s) ↔ NH3(g) + H2S(g)
a) Give the expressions for Kc and Kp
Kc = [NH3(g)][H2S(g)]
Kp = (PNH3)(PH2S)
b) When the system reaches equilibrium determine the concentration of NH3 and H2S.
Kc = [NH3(g)][H2S(g)] Î 1.8*10-4 M2 = x2 Î x= 0.013 M
Let the concentrations of both gases be equal to x.
c) Determine the equilibrium concentrations if an initial concentration of NH3 of 0.020 M
were present.
NH4HS
NH3
H2S
Initial
-
0.020
0
∆
0
x
x
Final
-
0.020 + x
x
Kc = (0.020 M + x) (x) Î 0 = x2 + 0.020 x –1.8*10-4
Use the quadratic formula Î x = 0.0067 M
Therefore [NH3] = 0.020 M + 0.0067 M = 0.027 M and [H2S] = 0.0067 M
d) Determine the concentrations of the NH3 and H2S when the system reaches
equilibrium after the initial volume is halved.
C = n/(V/2) = 2(n/V) = 2C Î the concentrations will initially double and then
equilibrium will be reestablished.
NH4HS
NH3
H2S
Initial
-
0.0268 M
0.0268 M
∆
0
-x
-x
Final
-
0.0268 - x
0.0268 - x
Kc = (0.0268 –x)2 Î 0 = x2 –0.0536x –5.38*10-4
Use the quadratic equations and x is found to be 0.013 M