Final

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:
 

Freezing point depression due to CaCl₂:
 

CaCl₂ = Ca²⁺ + 2 Cl^- ⇒ i = 3

Molality: m (in mol.kg^-1) = n_CaCl2 (in mol) / m_water (in kg)

n_CaCl2 = M_CaCl2 x V_solution

and m_water = d_water x V_solution with d_water = volumetric mass density of water = 1.00 kg.L^-1
 

⇒ m = M_CaCl2 / d_water = 2.00 mol.kg^-1

Freezing point depression:

ΔT_f = - i x K_f x m = - 3 x 1.853 x 2.00 = -11.1 K = - 11.1°C

⇒ T_{f solution} – T_{f water} =  - 11.1°C

⇒ T_{f solution} – 0 =  - 11.1°C

⇒ T_{f solution} = - 11.1°C

Rate = k [NO]^α [O₂]^β

Let’s determine α:

[NO]_exp2 = 2 x [NO]_exp1; [O₂]_exp2 = [O₂]_exp1 and rate_exp2 = 2 x rate_exp1

rate_exp2 = 2 x rate_exp1

⇒ k [NO]_exp2^α [O₂]_exp2^β = 2 k [NO]_exp1^α [O₂]_exp1^β

⇒ k 2^α x [NO]_exp1 ^α [O₂]_exp1^β = 2 k [NO]_exp1^α [Cl₂]_exp1^β

⇒ 2^α = 2

⇒ α = 1

Let’s determine β:

[NO]_exp3 = [NO]_exp2; [O₂]_exp3 = 2 [O₂]_exp2 and rate_exp3 = 4 x rate_exp2

rate_exp3 = 4 x rate_exp2

⇒ k [NO]_exp3^α [O₂]_exp3^β = 4 k [NO]_exp2^α [O₂]_exp2^β

⇒ k [NO]_exp2 ^α 2 ^β [O₂]_exp2^β = 4 k [NO]_exp2^α [O₂]_exp2^β

⇒ 2 ^β = 4

⇒ β = 2

Rate = k [NO][O₂]²

Rate_exp1 = k [NO]_exp1 [O₂]_exp1²

⇒ k = $\frac{{\mathrm{Rate}}_{\exp 1}}{{\left[\mathrm{NO}\right]}_{\exp 1} {[\mathrm{O}}^{}}$₂]_exp12 = $\frac{0.032}{0.010 \times 0.{015}^2}$ = 1.4 x 10⁴ M ^-2.s^-1

1) 2 NO + O₂ → 2 NO₂

2) Formation of nitrogen: - $\frac{∆\left[{\mathrm{NO}}_2\right]}{∆\mathrm{t}}$

rate law: - $\frac{1}{2} \frac{∆\left[{\mathrm{NO}}_2\right]}{∆\mathrm{t}}$ = 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

- $\frac{1}{2} \frac{∆\left[{\mathrm{NO}}_2\right]}{∆\mathrm{t}}$ = k₂[N₂O₂][O₂]

⇒ - $\frac{∆\left[{\mathrm{NO}}_2\right]}{∆\mathrm{t}}$ = 2 k₂[N₂O₂][O₂]

⇒ - $\frac{∆\left[{\mathrm{NO}}_2\right]}{∆\mathrm{t}}$ = 2 $\frac{{\mathrm{k}}_2 {\mathrm{k}}_1}{{\mathrm{k}}_{-1}}$[NO]²[O₂] 

The rate law for the formation of nitrogen dioxide is 2 $\frac{{\mathrm{k}}_2 {\mathrm{k}}_1}{{\mathrm{k}}_{-1}}$  [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] = $\frac{{\mathrm{k}}_1 {\left[\mathrm{NO}\right]}^2}{{\mathrm{k}}_{-1} + {\mathrm{k}}_2 \left[{\mathrm{O}}_2\right]}$ 

- $\frac{∆\left[{\mathrm{NO}}_2\right]}{∆\mathrm{t}}$ = 2 k₂[N₂O₂][O₂]

⇒ - $\frac{∆\left[{\mathrm{NO}}_2\right]}{∆\mathrm{t}}$ = 2 $\frac{{\mathrm{k}}_2 {\mathrm{k}}_1 {\left[\mathrm{NO}\right]}^2 \left[{\mathrm{O}}_2\right]}{{\mathrm{k}}_{-1} + {\mathrm{k}}_2 \left[{\mathrm{O}}_2\right]}$

The rate law for the formation of nitrogen dioxide is 2 

Radioactive decay is a first-order process:

ln $\left(\frac{\mathrm{A}}{{\mathrm{A}}_0}\right)$ = - kt and t_1/2 = $\frac{\ln 2}{\mathrm{k}}$

$\frac{\mathrm{A}}{{\mathrm{A}}_0}$ = e^-kt = exp $\left(- \frac{\ln 2 \times \mathrm{t}}{{\mathrm{t}}_{1/2}}\right)$

A = 5.21 x 10¹² x exp $\left(- \frac{\ln 2 \times 135}{28.79}\right)$

A = 2.02 x 10¹¹ Becquerel

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

ΔH_rxn = Σ H_{bonds broken} – Σ H_{bonds formed} = Σ H_bond (reactants) - Σ H_bond (products)

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

Σ H_bond (reactants) < Σ H_bond (products)

⇒ ΔH_rxn < 0

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

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

n_KOH = 2 n_H2SO4

⇒ n_KOH = 2 x [H₂SO₄] x V_H2SO4 = 2 x 100.0 x 10^-3 x 0.275

⇒ n_KOH = 5.5 x 10^-2 mol

1)

E = $\frac{6.626 \times {10}^{-34} \times 2.998 \times {10}^8}{3.663 \times {10}^{-7}}$

E = 5.423 x 10^-19 J

2)  

E_{n moles photons} = n x N_A x E_{1 photon}

E_{n moles photons} = 1.50 x 10^-1 x 6.022 x 10²³ x 5.423 x 10^-19 J

E_{n moles photons} = 4.899 x 10⁴ J

First half reaction:

S₂O₃^2- → HSO₄^-   (unbalanced)

S₂O₃^2- + 5 H₂O → 2 HSO₄^- + 8 H⁺ + 8 e^-   (balanced)

Second half reaction:

Cl₂ → Cl^-   (unbalanced)

Cl₂ + 2 e^- → 2 Cl^-

Oxidation-reduction reaction:

S₂O₃^2- + 5 H₂O + 4 Cl₂ → 2 HSO₄^- + 8 H⁺ + 8 Cl^-