Extra Credit 8

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Q16.3.2

Determine the entropy change for the combustion of liquid ethanol, C₂H₅OH, under standard state conditions to give gaseous carbon dioxide and liquid water.

Challenge: Find the equilibrium constant for this reaction at 298.15 K, using state variables.

Solution

We ought to know what is happening:

\[ \ce{C2H5OH (l)} + 3\ce{O2 (g)} \rightarrow 2\ce{CO2 (g)} + 3\ce{H2O (l)} \]

Since we're working with standard state variables, we can consult a reference for our values of entropy: Table T1.

From which we find:

$$ Molecule $$	\[ S° \frac{J}{mol \ K} \]
$$ \ce{C2H5OH (l)} $$	$$ 160.7 $$
$$ \ce{O2 (g)} $$	$$ 205.2 $$
$$ \ce{CO2 (g)} $$	$$ 213.8 $$
$$ \ce{H2O (l)} $$	$$ 109.6 $$

Then we may apply Hess's Law to obtain: \[ \Delta S = 3 \cdot S_{\ce{CO2 (g)}} + 3 \cdot S_{\ce{H2O (l)}} - 2 \cdot S_{\ce{O2 (g)}} - S_{\ce{C2H5OH (l)}} \]

%Input answer...

Answer: $\Delta S = 399.1 $

Q5.4.3

Calculate ΔH for the reaction described by the equation.

$\ce{Ba(OH)2⋅8H2O}(s)+\ce{2NH4SCN}(aq)⟶\ce{Ba(SCN)2}(aq)+\ce{2NH3}(aq)+\ce{10H2O}(l)$

Solution:

We need only apply Hess's Law:

Q10.3.9

Carbon tetrachloride, CCl₄, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °C, the vapor pressure of CCl₄ is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl₄.

Solution:

We know the enthalpy of vaporization of our molecule, and two observations: Temperature, and an associated Vapor Pressure. We want to know what temperature will induce a new Vapor Pressure (1 atm).

This is enough to apply the Clausius-Clapeyron equation.

\[ ln(\frac{P_1}{P_2}) = -\frac{\Delta H_{vap}}{R}[\frac{1}{T_1} - \frac{1}{T_2}] \]

%input answer

Answer: $\ T = 349.35K = 76.1^o C $

Q13.2.23

Write the expression of the reaction quotient for the ionization of HOCN in water.

Solution:

Recall that the reaction quotient is the activity of the products, divided by that of the reactants.

Find the reaction: \[ \ce{HOCN} + \ce{H2O} \rightleftharpoons \ce{H3O}^+ + \ce{OCN}^- \]

The rest is straightforward:

%input answer

Answer:

\[ Q = \frac{[H_3O^+][OCN]}{[H_2O][HOCN]} \]

Q14.2.2

The ionization constant for water (K_w) is 2.9 × 10⁻¹⁴ at 40 °C. Calculate [H₃O⁺], [OH⁻], pH, and pOH for pure water at 40 °C.

Solution:

Step 1: Find the reaction: \[\ce{H2O} \rightleftharpoons \ce{H}^+ + \ce{OH}^- \]

Step 2: Find the equilibrium expressions

We can justly approximate K_was the product of concentrations \[ \K_w = \ce{[H3O^+]} \cdot \ce{[OH^-]} \]

And therefore each concentration is the square root of K_w.

Step 3: Find the pH and pOH.

We know that \[pH \approx -log\ce{[H3O^+]} \], and likewise for \ce{OH^-}. We need only substitute each. Note: we are not at standard conditions, so we cannot use the typical pH + pOH = 14 relation.

%input answer

Answer:

$\ [H_3O^+] = [OH^-] = 1.703 x 10^{-7} $

$\ pH = pOH = 6.77 $

Q15.1.X

The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate K_sp for each of the slightly soluble solids indicated:

AgBr: [Ag⁺] = 5.7 × 10^–7 M, [Br^–] = 5.7 × 10^–7 M
CaCO₃: [Ca²⁺] = 5.3 × 10^–3 M, $\ce{[CO3^2- ]}$ = 9.0 × 10^–7 M
PbF₂: [Pb²⁺] = 2.1 × 10^–3 M, [F^–] = 4.2 × 10^–3 M
Ag₂CrO₄: [Ag⁺] = 5.3 × 10^–5 M, 3.2 × 10^–3 M
InF₃: [In³⁺] = 2.3 × 10^–3 M, [F^–] = 7.0 × 10^–3 M

Solution:

Answer:

a. $\ K_{sp} = 3.249 x 10^{-13} $

b. $\ K_{sp} = 4.77 x 10^{-9} $

c. $\ K_{sp} = 3.704 x 10^{-8} $

d. $\ K_{sp} = 8.989 x 10^{-12} $

e. $\ K_{sp} = 7.889 x 10^{-10} $