# Midterm 1

General Chemistry 3

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Here is an equilibrium: C (s) + CO2 (g) $⇌$ 2 CO (g)     [ΔH0rxn = 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 N2 (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 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

[SO3]0 = 2.00 M, [O2]0 = 0.0850 M, [SO2]0 = 0 M and [SO2]eq = 0.120 M

Calculate Kc for the following reaction:

2 SO3 (g) $⇌$ 2 SO2 (g) + O2 (g)

Kc =

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

Kc =

Kc =

Kc = 5.91 x 10-4

Calculate the pH of a 0.100 M aqueous solution of ethylamine CH3CH2NH2.

Data: Kb [CH3CH2NH2] = 4.30 x 10-4

Acido-basic reaction:

CH3CH2NH2 (aq) + H2O (l) $⇌$ CH3CH2NH3+ (aq) + HO- (aq)     [Kb]

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

Kb = $\frac{\left[{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}{{\mathrm{NH}}_{3}}^{+}\right]\left[{\mathrm{HO}}^{-}\right]}{\left[{\mathrm{CH}}_{3}{\mathrm{CH}}_{2}{\mathrm{NH}}_{2}\right]}$  =

Let’s suppose 0.100 >> X

Kb = 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 Kb for this weak base.

A- + H2O $⇌$ AH + HO-     [Kb]

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 = 10pH – 14 = 10-2.7

⇒ [HO-] = 2.00 x 10-3 M

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

K=

Kb = 7.23 x 10-4

Write the relative reaction rate equation for each component in the combustion of ethanol CH3CH2OH.

Combustion of ethanol:

CH3CH2OH + O2 → CO2 + H2O     [unbalanced equation]

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

Reaction rate:

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