# Midterm 1

General Chemistry 3

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Here is an equilibrium: C (s) + CO_{2} (g) $\rightleftharpoons $ 2 CO (g) [ΔH^{0}_{rxn} = 65 kJ]

Predict and explain the direction that the equilibrium will shift when each of the following changes occurs:

- CO is removed at constant temperature

- the total pressure is increased by addition of N_{2} (g) at constant temperature

- the volume is increased at constant temperature

- the temperature is decreased

- 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 N_{2} (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

[SO_{3}]_{0} = 2.00 *M*, [O_{2}]_{0} = 0.0850 *M*, [SO_{2}]_{0} = 0 *M* and [SO_{2}]_{eq} = 0.120 *M*

Calculate K_{c} for the following reaction:

2 SO_{3} (g) $\rightleftharpoons $ 2 SO_{2} (g) + O_{2} (g)

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:

[SO_{3}]_{eq} = [SO_{3}]_{0} – 2X

[O_{2}]_{eq} = [O_{2}]_{0} + X

[SO_{2}]_{eq} = [SO_{2}]_{0} + 2X = 2X = 0.120 *M*

⇒ X = 6.00 x 10^{-2} *M*

At the equilibrium:

[SO_{3}]_{eq} = [SO_{3}]_{0} – 2X = 2.00 – 0.120 = 1.88 *M*

[O_{2}]_{eq} = [O_{2}]_{0} + X = 0.0850 + 0.0600 = 0.145 *M*

[SO_{2}]_{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}

Calculate the pH of a 0.100 *M* aqueous solution of ethylamine CH_{3}CH_{2}NH_{2}.

Data: K_{b} [CH_{3}CH_{2}NH_{2}] = 4.30 x 10^{-4}

Acido-basic reaction:

CH_{3}CH_{2}NH_{2} (aq) + H_{2}O (l) $\rightleftharpoons $ CH_{3}CH_{2}NH_{3}^{+} (aq) + HO^{-} (aq) [K_{b}]

A t = 0:

[CH_{3}CH_{2}NH_{2}] = 0.100 *M* and [CH_{3}CH_{2}NH_{3}^{+}] = [HO^{-}] = 0 *M*

At the equilibrium:

[CH_{3}CH_{2}NH_{2}] = 0.100 - X *M* and [CH_{3}CH_{2}NH_{3}^{+}] = [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 7.50 x 10^{-3} *M* solution of a weak base has a pH of 11.3.

Determine the K_{b} for this weak base.

A^{-} + H_{2}O $\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}

Write the relative reaction rate equation for each component in the combustion of ethanol CH_{3}CH_{2}OH.

Combustion of ethanol:

CH_{3}CH_{2}OH + O_{2 }→ CO_{2} + H_{2}O [unbalanced equation]

CH_{3}CH_{2}OH + 3 O_{2 }→ 2 CO_{2} + 3 H_{2}O [balanced equation]

Reaction rate:

rate = - $\frac{\u2206\left[{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{OH}\right]}{\u2206\mathrm{t}}$ = - $\frac{1}{3}$ $\frac{\u2206\left[{\mathrm{O}}_{2}\right]}{\u2206\mathrm{t}}$ = $\frac{1}{2}$ $\frac{\u2206\left[{\mathrm{CO}}_{2}\right]}{\u2206\mathrm{t}}$ = $\frac{1}{3}$ $\frac{\u2206\left[{\mathrm{H}}_{2}\mathrm{O}\right]}{\u2206\mathrm{t}}$