Problem Set Answers (2012-07-02)

Chem 210 - Summer 2012 - Problem Set {2012-07-02}

1. A 395.4g block of glass (heat capacity = 0.84J/g.K) warms from -42.37℃ to 20.53℃. How much energy/heat has been transferred? Is this process endothermic or exothermic? If the energy/heat gained or lost by the glass block is transferred from/to a 528.7g block of copper (heat capacity = 0.385J/g.K) initially at 15.39℃, what is the final temperature of the copper block?

(0.84J/g•K)(395.4g)(62.90K)  =  20891J  →  20.9kJ

If the glass block is warming, then energy/heat is being absorbed by the block, this is an endothermic process.

(0.385J/g•K)(528.7g)(ΔT)  =  20891J

ΔT  =  102.6°C

Since the glass block is absorbing heat (endothermic), the copper block must be releasing heat and getting colder.

15.39°C – 102.6°C  =  -87.2°C

2. You have dissolved 12.482g of aluminum nitrate in enough water to make 150.0mL of solution. What is the molarity of aluminum ions in the resulting solution? What is the molarity of nitrate ions in the resulting solution?
First, find the molarity of the whole aluminum nitrate

(12.482g) / (212.994g/mol)  =  0.0586026mols Al(NO₃)₃

(0.0586026mols Al(NO₃)₃) / (0.1500L)  =  0.390684M Al(NO₃)₃ (aq)

For every mol of Al(NO₃)₃ that dissolves, there is 1 mol of Al⁺³(aq). Converting…

(0.390684mols Al(NO₃)₃ / L) (1mol Al⁺³(aq) / mols Al(NO₃)₃)  =  0.3907M Al⁺³(aq)

For every mol of Al(NO₃)₃ that dissolves, there are 3 mol of NO₃^-1(aq). Converting…

(0.390684mols Al(NO₃)₃ / L) (3mol NO₃^-1(aq) / mols Al(NO₃)₃)  =  1.1721M NO₃^-1(aq)

3. What is the freezing point of a solution made from 7.113g of potassium sulfate dissolved in 75.00mL of water?
Calculate the molality of potassium sulfate:

(7.113g) / (174.258g/mol)  =  0.040819mols K₂SO₄

(0.040819mols K₂SO₄) / (0.07500kg)  =  0.54425m K₂SO₄(aq)

Plugging in to the freezing point depression equation:

ΔT_fp = k_fpd • m • i

ΔT_fp  =  (1.86°C/m)(0.54425m)(3)  =  3.04°C

The freezing point of the solution is depressed by 3.04°C. Since the freezing point of pure water is 0°C, the freezing point of the solution should be -3.04°C.  

4. Ammonia gas reacts with chlorine gas to form nitrogen trichloride gas and hydrogen gas. At some point, 53.65mg of nitrogen trichloride is formed in 28.63 seconds in a 12.00L reaction vessel. What is the rate of the reaction? What are the rates of consumption and production for each reactant and product in the mixture?
The only thing we have any information about is NCl₃, so let’s start by calculating the rate of formation of NCl₃ …

Rate_NCl3  =  { (0.05365g NCl₃ / 120.366g/mol) / (12.00L) } / 28.63sec  =  1.297x10^-6 M/sec

Now we need the balanced chemical equation:

2 NH₃(g)  +  3 Cl₂(g)  →  2 NCl₃(g)  +  3 H₂(g)

Rate_rxn should be half the rate of production of NCl₃ , so (1.297x10^-6 M/sec) / 2  =  6.487x10^-7 M/sec

Rate_NH3 should be the same as the rate of production of NCl₃ or 2xRate_rxn, 1.297x10^-6 M/sec

Rate_Cl2 should be three times Rate_rxn, 3(6.487x10^-7 M/sec) = 1.946x10^-6 M/sec

Rate_H2 should be three times Rate_rxn, 3(6.487x10^-7 M/sec) = 1.946x10^-6 M/sec

5. You are studying the reaction of sulfur dioxide gas with hydrogen chloride gas to form thionyl chloride {SOCl₂(g)} and water gas at 6.48℃. You've completed the following reactions:

Rxn #	[sulfur dioxide]0	[hydrogen chloride]0	Initial Rate (M/min)
1	0.429	0.317	3.02x10-5
2	0.858	0.317	1.21x10-4
3	0.429	0.951	9.06x10-5

What is the rate law expression for this reaction? Include the rate law constant. If Rxn #3 is repeated at 28.81℃, the rate increases to 4.67x10^-3 M/min. What is the activation energy for this process?
Comparing Rxn #1 and #2, doubling [SO₂]₀ quadruples the initial rate  →  2nd order with respect to [SO₂]₀
Comparing Rxn #1 and #3, tripling [HCl]₀ triples the initial rate  →  1st order with respect to [HCl]₀
The rate law expression is:

Rate₀  =  k [SO₂]₀² [HCl]₀¹

Plugging in conditions for Rxn #1 (or any of them…) to get the value of “k”:

3.02x10^-5 M/min  =  k (0.429M)² (0.317M)

k  =  5.176x10^-4 M^-2min^-1

For the new temperature, we need a new value of “k”, plugging in we get…

4.67x10^-3 M/min  =  k (0.429M)² (0.951M)

k  =  2.668x10^-2 M^-2min^-1

Plugging in to the comparative form of the Arrhenius equation:

ln (k₁/k₂)  =  (E_a/R) {(1/T₂) – (1/T₁)}

ln (5.176x10^-4 M^-2min^-1 / 2.668x10^-2 M^-2min^-1)  =  (E_a / 8.314J/mol•K) {(1/301.96K) – (1/279.63K)}

E_a  =  123.9kJ/mol

Any questions? Let me know.