# Final

General Chemistry 2

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Circle the molecule(s) that are likely to be water soluble:

Substances with similar types of intermolecular forces dissolve in each other.

Water is polar and exhibits hydrogen bonding

⇒ polar molecules and molecules exhibiting hydrogen bonding are likely to be water soluble:

Calculate the freezing point of a 2.00 M solution of CaCl2 in water.

Data: freezing point depression constant of water = 1.853 K.kg.mol-1

Freezing point depression due to CaCl2:

ΔTf = - i x Kf x m

ΔTf = Tf solutionCaCl2 – Tf water

i = van’t Hoff i-factor = number of ions/molecules produced per formula unit

m = solution molality (in mol.kg-1)

Kf = proportionality constant (freezing point depression constant)

CaCl2 = Ca2+ + 2 Cl- ⇒ i = 3

Molality: m (in mol.kg-1) = nCaCl2 (in mol) / mwater (in kg)

nCaCl2 = MCaCl2 x Vsolution

and mwater = dwater x Vsolution with dwater = volumetric mass density of water = 1.00 kg.L-1

⇒ m = MCaCl2 / dwater = 2.00 mol.kg-1

Freezing point depression:

ΔTf = - i x Kf x m = - 3 x 1.853 x 2.00 = -11.1 K = - 11.1°C

⇒ Tf solution – Tf water =  - 11.1°C

⇒ Tf solution – 0 =  - 11.1°C

⇒ Tf solution = - 11.1°C

Here is a reaction: 2 NO (g) + O(g) → 2 NO(g)

We measure the initial rate of reaction of 3 experiments with different initial concentrations.

Determine the rate constant of the reaction.

Exp1: [NO] = 0.010 M; [O2] = 0.015 M; rate = 0.032 M.s-1

Exp2: [NO] = 0.020 M; [O2] = 0.015 M; rate = 0.064 M.s-1

Exp3: [NO] = 0.010 M; [O2] = 0.030 M; rate = 0.128 M.s-1

Rate = k [NO]α [O2]β

Let’s determine α:

[NO]exp2 = 2 x [NO]exp1; [O2]exp2 = [O2]exp1 and rateexp2 = 2 x rateexp1

rateexp2 = 2 x rateexp1

⇒ k [NO]exp2α [O2]exp2β = 2 k [NO]exp1α [O2]exp1β

⇒ k 2α x [NO]exp1 α [O2]exp1β = 2 k [NO]exp1α [Cl2]exp1β

⇒ 2α = 2

⇒ α = 1

Let’s determine β:

[NO]exp3 = [NO]exp2; [O2]exp3 = 2 [O2]exp2 and rateexp3 = 4 x rateexp2

rateexp3 = 4 x rateexp2

⇒ k [NO]exp3α [O2]exp3β = 4 k [NO]exp2α [O2]exp2β

⇒ k [NO]exp2 α 2 β [O2]exp2β = 4 k [NO]exp2α [O2]exp2β

⇒ 2 β = 4

⇒ β = 2

Rate = k [NO][O2]2

Rateexp1 = k [NO]exp1 [O2]exp12

⇒ k = 2]exp12 =  = 1.4 x 104 -2.s-1

Nitric oxide NO reacts with dioxygen to form nitrogen dioxide.

The mechanistic process is:

Step 1: NO + NO $⇌$ N2O2 (k1 and k-1)

Step 2: N2O2 + O2 → 2 NO2 (k2)

1) Write a chemical equation for this reaction

2) Write the rate law for the formation of nitrogen dioxide using the mechanistic proposal assuming the first step is a fast equilibrium

3) Write the rate law for the formation of nitrogen dioxide without assumptions about fast or slow steps

Be sure to eliminate intermediates from the rate expression

1) 2 NO + O2 → 2 NO2

2) Formation of nitrogen: - $\frac{∆\left[{\mathrm{NO}}_{2}\right]}{∆\mathrm{t}}$

rate law: -  = k2[N2O2][O2]

N2O2 is not a reactant or a product: it should be eliminated from the rate law.

Step 1 is a fast equilibrium: k1[NO]2 = k-1 [N2O2]

⇒ [N2O2] = $\frac{{\mathrm{k}}_{1}}{{\mathrm{k}}_{-1}}$ [NO]2

= k2[N2O2][O2]

⇒ - $\frac{∆\left[{\mathrm{NO}}_{2}\right]}{∆\mathrm{t}}$ = 2 k2[N2O2][O2]

⇒ - $\frac{∆\left[{\mathrm{NO}}_{2}\right]}{∆\mathrm{t}}$ = 2 [NO]2[O2

The rate law for the formation of nitrogen dioxide is 2   [NO]2[O2

3) N2O2 is an intermediate:

k1[NO]2 – k-1[N2O2] – k2 [N2O2][O2] = 0

⇒ [N2O2] (k-1 + k2[O2]) = k1[NO]2

⇒ [N2O2] =

- $\frac{∆\left[{\mathrm{NO}}_{2}\right]}{∆\mathrm{t}}$ = 2 k2[N2O2][O2]

⇒ - $\frac{∆\left[{\mathrm{NO}}_{2}\right]}{∆\mathrm{t}}$ = 2

The rate law for the formation of nitrogen dioxide is 2

Strontium-90 has a specific activity of 5.21 x 1012 Becquerel and a half-time of 28.79 years.

Calculate the activity of 90Sr after 135 years have passed.

Radioactive decay is a first-order process:

ln $\left(\frac{\mathrm{A}}{{\mathrm{A}}_{0}}\right)$ = - kt and t1/2

$\frac{\mathrm{A}}{{\mathrm{A}}_{0}}$ = e-kt = exp

A = 5.21 x 1012 x exp

A = 2.02 x 1011 Becquerel

1) Draw the most stable Lewis structure for SF4 including lone pair and formal charge(s). Draw any resonance structures if appropriate.

2) What is the geometry of the molecule?

3) What is/are the F-S-F angle(s)?

1)

2) See-saw geometry

3) Because of the lone pair of electrons on the sulfur: the 2 F-S-F angles are < 90° and < 120°C

What is the sign of ΔHrxn0 for a reaction in which the bonds are stronger in the products than in the reactants? Explain.

ΔHrxn = Σ Hbonds broken – Σ Hbonds formed = Σ Hbond (reactants) - Σ Hbond (products)

If which the bonds are stronger in the products than in the reactants:

Σ Hbond (reactants) < Σ Hbond (products)

⇒ ΔHrxn < 0

If 100.0 mL of a 0.275 M sulfuric acid solution is needed to neutralize a solution of KOH, how many moles of KOH must be present in the solution?

Neutralization reaction: H2SO4 + 2 KOH → 2 H2O + K2SO4

According to the equation: 2 moles of KOH react with 1 mole of H2SO4

nKOH = 2 nH2SO4

nKOH = 2 x [H2SO4] x VH2SO4 = 2 x 100.0 x 10-3 x 0.275

⇒ nKOH = 5.5 x 10-2 mol

A new prototype UV light source emits photons at a wavelength of 366.3 nm

1) Calculate the energy per photon.

2) Calculate the total energy associated with the emission of 1.50 x 10-1 moles of photons at this wavelength

1)

E = $\frac{\mathrm{hc}}{\mathrm{\lambda }}$

h = Planck’s constant = 6.626 x 10-34 m2.kg.s-1

c = speed of light = 2.998 x 108 m.s-1

λ = wavelength (in m) à 366.3 nm = 3.663 x 10-7 m

E =

E = 5.423 x 10-19 J

2)

En moles photons = n x NA x E1 photon

n = number of moles (in mol)

NA = Avogadro’s number = 6.022 x 1023

En moles photons = n x NA x E1 photon

En moles photons = 1.50 x 10-1 x 6.022 x 1023 x 5.423 x 10-19 J

En moles photons = 4.899 x 104 J

Thiosulfate anion (S2O32-) is used in fiber industry for the treatment of excess chlorine.

Balance the oxidation-reduction reaction under acidic conditions knowing that the weak acid HSO4- is formed.

First half reaction:

S2O32- → HSO4-   (unbalanced)

S2O32- + 5 H2O → 2 HSO4- + 8 H+ + 8 e-   (balanced)

Second half reaction:

Cl2 → Cl-   (unbalanced)

Cl2 + 2 e- → 2 Cl-

Oxidation-reduction reaction:

S2O32- + 5 H2O + 4 Cl2 → 2 HSO4- + 8 H+ + 8 Cl-