Extra Credit 8
- Page ID
- 96939
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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 (Kw) is 2.9
Solution:
Step 1: Find the reaction: \[\ce{H2O} \rightleftharpoons \ce{H}^+ + \ce{OH}^- \]
Step 2: Find the equilibrium expressions
We can justly approximate Kw as the product of concentrations \[ \K_w = \ce{[H3O^+]} \cdot \ce{[OH^-]} \]
And therefore each concentration is the square root of Kw.
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 Ksp 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. \(\ K_{sp} = 7.889 x 10^{-10} \)