Midterm 1

- CO is removed at constant temperature:

direction: toward product

explanation: when product is removed, the reaction shifts to make more product (Le Chatelier’s principle)

- the total pressure is increased by addition of N₂ (g) at constant temperature

direction: no change

explanation: volume did not change à partial pressure did not change à Q did not change à no shift

- the volume is increased at constant temperature

direction: toward product

explanation: side with most moles of gas

- the temperature is decreased

direction: toward reactants

explanation: reaction is endothermic

K_c = $\frac{{{\left[{\mathrm{SO}}_2\right]}_{\mathrm{eq}}}^2 {\left[{\mathrm{O}}_2\right]}_{\mathrm{eq}}}{{{\left[{\mathrm{SO}}_3\right]}_{\mathrm{eq}}}^2}$

From the stoichiometry of the reaction:

[SO₃]_eq = [SO₃]₀ – 2X

[O₂]_eq = [O₂]₀ + X

[SO₂]_eq = [SO₂]₀ + 2X = 2X = 0.120 M

⇒ X = 6.00 x 10^-2 M

At the equilibrium:

[SO₃]_eq = [SO₃]₀ – 2X = 2.00 – 0.120 = 1.88 M

[O₂]_eq = [O₂]₀ + X = 0.0850 + 0.0600 = 0.145 M

[SO₂]_eq = 0.120 M

K_c = $\frac{{{\left[{\mathrm{SO}}_2\right]}_{\mathrm{eq}}}^2 {\left[{\mathrm{O}}_2\right]}_{\mathrm{eq}}}{{{\left[{\mathrm{SO}}_3\right]}_{\mathrm{eq}}}^2}$

K_c = $\frac{0.{120}^2 \times 0.145}{1.{88}^2}$

K_c = 5.91 x 10^-4

Acido-basic reaction:

CH₃CH₂NH₂ (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂NH₃⁺ (aq) + HO^- (aq)     [K_b]

A t = 0:

[CH₃CH₂NH₂] = 0.100 M and [CH₃CH₂NH₃⁺] = [HO^-] = 0 M

At the equilibrium:

 [CH₃CH₂NH₂] = 0.100 - X M and [CH₃CH₂NH₃⁺] = [HO^-] = X M

K_b = $\frac{\left[{\mathrm{CH}}_3{\mathrm{CH}}_2{{\mathrm{NH}}_3}^{+}\right]\left[{\mathrm{HO}}^{-}\right]}{\left[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2\right]}$  =  $\frac{{\mathrm{X}}^2}{0.100 - \mathrm{X}}$

Let’s suppose 0.100 >> X

K_b = 4.30 x 10^-4 = $\frac{{\mathrm{X}}^2}{0.100}$

⇒ X = 6.56 x 10^-3 (X cannot be < 0)

⇒ the hypothesis 0.100 >> X is verified

pOH = - log [HO^-] = 2.18

pH = 14 – pOH = 11.8

A^- + H₂O $\rightleftharpoons$ AH + HO^-     [K_b]

A t = 0:

[A^-] = 7.50 x 10^-3 M and [AH] = [HO^-] = 0 M

At the equilibrium:

[A^-] = 7.50 x 10^-3 - X M and [AH] = [HO^-] = X M

pH = 14 – pOH = 14 + log [HO^-]

⇒ [HO^-] = X = 10^{pH – 14} = 10^-2.7

⇒ [HO^-] = 2.00 x 10^-3 M

K_b = $\frac{\left[\mathrm{AH}\right]\left[{\mathrm{HO}}^{-}\right]}{\left[{\mathrm{A}}^{-}\right]}$ 

K_b = $\frac{{\left(2.00 \times {10}^{-3}\right)}^2}{7.50 \times {10}^{-3} - 2.00 \times {10}^{-3}}$

K_b = 7.23 x 10^-4

Combustion of ethanol:

CH₃CH₂OH + O₂ → CO₂ + H₂O     [unbalanced equation]

CH₃CH₂OH + 3 O₂ → 2 CO₂ + 3 H₂O     [balanced equation]

Reaction rate:

rate = - $\frac{∆\left[{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\right]}{∆\mathrm{t}}$ = - $\frac{1}{3}$ $\frac{∆\left[{\mathrm{O}}_2\right]}{∆\mathrm{t}}$ = $\frac{1}{2}$ $\frac{∆\left[{\mathrm{CO}}_2\right]}{∆\mathrm{t}}$ = $\frac{1}{3}$ $\frac{∆\left[{\mathrm{H}}_2\mathrm{O}\right]}{∆\mathrm{t}}$