SCH4U - Energy Changes and Rates of Reactions Sample Test Qs

hi guyz it is exam season! i dont feel like typing a real post so im just gonna do some sample problems hehehe

1. Sehee used a coffee cup calorimeter to dissolve 11.86 g of NH4NO3 in 100 mL of water. The initial temperature of the water was 20.5 degC. At the end of the reaction, the temperature recorded was 11.6 degC. Calculate the enthalpy change in kJ/mol of NH4NO3. Indicate if the reaction is exothermic or endothermic. (Heat capacity of water = 4.18 J/g degC) (5 marks)

Answer:

NH4NO3	Water
ΔHsol'n = ? kJ/mol m = 11.86 g M = 2MN + 4MH + 3MO M = [2(14) + 4(1) + 3(16)] g/mol M = 80 g/mol n = m/M n = (11.86 g) / (80 g/mol) n = 0.1483 mol	m = 100 g T1 = 20.5 degC T2 = 11.6 degC ΔT = (20.5 - 11.6) degC ΔT = 8.9 degC

nΔHsol'n = mcΔT

nΔHsol'n = mcΔT / n

nΔHsol'n = (100 g)(4.18 J/g degC)(8.9 degC) / 0.1483

nΔHsol'n = 25086 J/mol (extra digits carried)

nΔHsol'n = 25.1 kJ/mol

the temperature increased, so the reaction is endothermic

2. The complete combustion of liquid acetic acid involves an enthalpy change of -876.1 kJ/mol.
a) Rewrite this as a balanced chemical equation, with the enthalpy change as part of the equation. (1 mark)

Answer:

CH3COOH(l) + 2 O2(g) → 2 H2O(l) + 2 CO2(g) + 876.1 kJ

b) Draw and label an enthalpy diagram for the above thermochemical reaction. (2 marks)

Answer:

c) How much heat is transferred when 147.0 g of acetic acid is combusted with excess oxygen according to the above equation? Is the reaction exothermic or endothermic? (4 marks)

Answer:

m = 147.0 g

M = 2MC + 4MH + 2MO

M = [2(12) + 4(1) + 2(16)] g/mol

M = 60 g/mol

n = m/M

n = (147.0 g) / (60 g/mol)

n = 2.45 mol

ΔH = nΔHcomb

ΔH = (2.45 mol)(-876.1 kJ/mol)

ΔH = -2146 kJ

ΔH is negative so it's exothermic

d) The molar enthalpies of formation for liquid water and carbon dioxide gas are -285.83 kJ/mol and -393.52 kJ/mol, respectively. Given these values, calculate the molar enthalpy of formation for acetic acid. (7 marks)

Answer:

Our given equations are:

(1): H2(g) + ½ O2(g) → H2O(l) + 285.83 kJ

(2): C(s) + O2(g) → CO2(g) + 393.52 kJ

(3): CH3COOH(l) + 2 O2(g) → 2 H2O(l) + 2 CO2(g) + 876.1 kJ

Our target equation is: H2(g) + C(s) + O2(g) → CH3COOH(l)

2 × (1): 2 H2(g) + O2(g) → 2 H2O(l) + 571.66 kJ

2 × (2): 2 C(s) + 2 O2(g) → 2 CO2(g) + 787.04 kJ

-1×(3): 2 H2O(l) + 2 CO2(g) + 876.1 kJ → CH3COOH(l) + 2 O2(g) 

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2 H2(g) + 2 C(s) + O2(g) → CH3COOH(l) + 482.6 kJ

4. The rate of the reaction: H2(g) + I2(g) ---> 2 HI(g)

Has been measured at the reactant concentrations shown (in mol/L)

Experiment	[H2]	[I2]	Initial rate (mol/(Ls))
1	0.010	0.050	0.0400
2	0.020	0.050	0.0800
3	0.030	0.050	0.120
4	0.050	0.010	0.0200
5	0.050	0.020	0.160
6	0.050	0.030	0.540

a) What is the order of reaction with respect to each of the reactants? (2 x 3 marks)

r = k[A]x[B]y

Change in [H2] = Trial 2/Trial 1

Change in [H2] = (0.020 mol/L)/(0.010 mol/L)

Change in [H2] = 2 

Change in rate = Trial 1/Trial 2 

Change in rate = (0.0800 mol/Ls)/(0.0400 mol/Ls)

Change in rate = 2

Change in [H2]x = Change in rate

2x = 2

 x = 1 

Change in [I2] = Trial 5/Trial 4

Change in [I2] =  (0.020 mol/L)/(0.010 mol/L)

Change in [I2] =  2

Change in rate = Trial 5/Trial 4

Change in rate = (0.160 mol/Ls)/(0.0200 mol/Ls)

Change in rate = 8

Change in [I2]y = Change in rate

2y = 8

y = 3

H2 is a first order reactant and I2 is a third order reactant.

b) What is the overall order of reaction? (1/2 mark)

1 + 3 = 4th order

c) Calculate a value for the rate constant (2 marks)

k = r/[H2][I2]3

k = (0.0400 mol/Ls) / [(0.010 mol/L)(0.050 mol/L)3]

k = 32 000 L3/(mol3s) 

d) What is the complete rate equation for the above equation?

r = 32000 L3/(mol3s)[H2][I2]3

e) According to these results, what would be the initial rate (in mol/(Ls)) if the concentrations of the reactants are: [H2]=0.200 mol/L and [I2]=1.60 mol/L? (1 mark)

r = 32000 L3/(mol3s)(0.200 mol/L)(1.60 mol/L)3

r = 26.2・104 mol/Ls