Extra Credit 25

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Q16.4.21

Consider the decomposition of CaCO₃(s) into CaO(s) and CO₂(g). What is the equilibrium partial pressure of CO₂ at room temperature?

The equation for this reaction is

$\ce{CaCO3_{(s)}} \leftrightharpoons \ce{CaO_{(s)}} + \ce{CO2_{(g)}}$

And the equilibrium constant is

$$ K_p = {\ce{[CO2]}} $$

To calculate the partial pressure of CO₂, K_p must first be calculated with the equation $\Delta G^\circ_{rxn} = -RT\ln(K_p)$. $\Delta G = 0$ for this reaction since it is at equilibrium. The $\Delta G^\circ_{rxn}$ and be calculated with the $\Delta G^\circ_{f}$ for each species

$\ce{CaCO3_{(s)}}: \Delta G^\circ_{f} = -1128.8 (kJ/mol)$

$\ce{CaO_{(s)}}: \Delta G^\circ_{f}$ = -604.2 (kJ/mol)\)

$\ce{CO2_{(g)}}: \Delta G^\circ_{f}$ = -394.6 (kJ/mol)\)

$$ \Delta G^\circ_{rxn} = {((-604.2) + (-394.6)) - (-1128.8)} $$

$$ \Delta G^\circ_{rxn} = {130 kJ/mol} $$

We know $T = 25^\circ C (298 K)$ and R = 8.314 J/mol*K as imputing these values into the equation above will determine the equilibrium constant

$$ \Delta G^\circ_{rxn} = {-RT\ln(K_p)} $$

$$ 1.3*10^{5} = {-(8.314)(298)\ln{K_p}} $$

$$ K_p = {1.63*10^{-23} = \ce{[CO2]}} $$

It makes sense that the partial pressure of the CO₂ would be so small since CaCO₃ does not readily dissociate so there would be a very small amount of CO₂ produced from this reaction.

Q15.1.X

Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Table E3 for K_sp values.)

KClO₄: [K⁺] = 0.01 M, $\ce{[ClO4- ]}$ = 0.01 M
K₂PtCl₆: [K⁺] = 0.01 M, $\ce{[PtCl6^2- ]}$ = 0.01 M
PbI₂: [Pb²⁺] = 0.003 M, [I^–] = 1.3 × 10^–3 M
Ag₂S: [Ag⁺] = 1 × 10^–10 M, [S^2–] = 1 × 10^–13 M

Solution:

For the compound to precipitate from the concentrations given, Q>K so that the reaction will shift to the left to form the precipitate. For each compound, Q much be calculated and compared to the K_sp value given in Table E3. Since each compound is a solid, The Equilibrium constant will be a product of the reactants.

a. $\ce{KClO^4_{(s)}} \leftrightharpoons {K^+_{(aq)}} + {ClO^{4-}_{(aq)}}$

$Q= \ce{[ClO4^{-}]}$$\ce{[K+]}$ $Q = (0.01)(.01) = 1.0*10^{-4}$

$K_{sp} = 1.05*10^{-2}$ $Q<K$

This solution would not produce the precipitate

b. $\ce{K_2PtCl_{6(s)}} \leftrightharpoons {2K^+_{(aq)}} + {PtCl^{2-}_{6(aq)}}$

$Q= \ce{[2K+]^2}$$\ce{[PtCl6^2- ]}$ $Q = (.01)^2(.01) = 1.0*10^{-6}$

$K_{sp} = 7.48*10^{-6}$ $Q<K$

This solution will not percipitate to form the compound

c. $\ce{PbI_{2(s)}} \leftrightharpoons {Pb^{2+}_{(aq)}} + {2I^-_{(aq)}}$

$Q= \ce{[Pb^{2+}]}$$\ce{[I^-]^2}$ $Q = (.003)(1.3*10^{-3})^2 = 5.1*10^{-9}$

$K_{sp} = 9.8*10^{-9}$ $Q<K$

This solution will not percipitate

d. $\ce{Ag_2S_{(s)}} \leftrightharpoons {2Ag^+_{(aq)}} + {S^{2-}_{(aq)}}$

$Q= \ce{[Ag+]^2}$$\ce{[S^2- ]}$ $Q = (1.0*10^{-10})^2(1.0*10^{-13}) = 1.0*10^{-33}$

$K_{sp} = 6.3*10^{-50}$ $Q>K$

Since K>Q, the reaction will shift to the left and precipitate to form the compound Ag₂S.

Q14.6.5

What is [H₃O⁺] in a solution of 0.075 M HNO₂ and 0.030 M NaNO₂?

$\ce{HNO2_{(aq)}}+\ce{H2O_{(l)}}⇌\ce{H3O+_{(aq)}}+\ce{NO2^-_{(aq)}} \hspace{20px} K_\ce{a}=4.5×10^{−5}$

Solution:

To find the concentration of hydronium ions for this buffer solution, an ICE table will be used to calculate the concentrations at equilibrium with the equilibrium constant given. HNO₂ is a weak acid so it will only partially dissociate, and NaNO₂ will completely dissociate into its ions.

	$\ce{[HNO2]}$	$\ce{[H3O+]}$	$\ce{[NO2^-]}$
Initial	$0.75 M$	$0$	$0.030M$
Change	$-x$	$+x$	$+x$
Equilibrium	$0.75-x$	$x$	$0.03+x$

The equilibrium constant equation is

$$K_a = {\ce{[H3O+]}\ce{[NO2^{-}]} \over \ce{[HNO2]}} $$

And with the K_aprovided and the equilibrium values from the ICE table, we get the equation

$$4.5*10^{-5} = {(x)(.03+x) \over (0.75 -x)} $$

When solved using a graphing calculator or the quadratic formula, the resulting solution is

$x=0.0011$

which when plugged back into the equilibrium value for the concentration of hydronium ions equals

$\ce{[H3O+]} = 0.0011$

Q13.3.13

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.

Write the expression for the equilibrium constant (K_c) for the reversible reaction \[\ce{Fe2O3}(s)+\ce{3H2}(g)\rightleftharpoons\ce{2Fe}(s)+\ce{3H2O}(g) \hspace{20px} ΔH=\mathrm{98.7\:kJ}\]
What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?
What will happen to the concentration of each reactant and product at equilibrium if H₂O is removed?
What will happen to the concentration of each reactant and product at equilibrium if H₂ is added?
What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?
What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

Solution:

a. $\ce{Fe_{(s)}}$ and $\ce{Fe2O3_{(s)}}$ will be left out of the equilibrium expression since they are solids. The equilibrium expression is as follows:

$$K_c = { \ce{[H2O]^3} \over \ce{[H2]^3} } $$

b. Since Fe is a solid adding more of it will not affect the equilibrium so there will be not change in the concentrations of the reactants or the products.

c. If H₂O is removed, Q will become less than K so reactants will decrease and products will increase to reestablish equilibrium.

d. If H₂ is added, Q will become less then K so the reactant concentrations will decrease and product concentrations will increase to reestablish equilibrium concentration.

e. If the pressure is increased, the reactants will be favored since there is 4 moles of reactants and 5 moles of products, so reactant concentrations will increase and product concentrations will decrease.

f. Since $ΔH=\mathrm{98.7\:kJ}$, the reaction is endothermic, so an increase in temperature would increase K and cause an increase in concentrations of reactants and a decrease in concentration of products.

Q16.4.25B

Acetic acid, CH₃CO₂H, can form a dimer, (CH₃CO₂H)₂, in the gas phase.

\[\ce{2CH3CO2H}(g)⟶\ce{(CH3CO2H)2}(g)\]

The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer.

At 25 °C, the equilibrium constant for the dimerization is 1.3 × 10³ (pressure in atm). What is ΔS° for the reaction?

To caculate the $\Delta S^\circ$ for the reaction, the equation

$$ \Delta G^\circ = {-RT\ln{K_p}} $$

can be substituted into the equation

$$ \Delta G^\circ = {\Delta H^\circ - T\Delta S^circ } $$

to get the equation which can be used to solve for $\Delta S^\circ$ for the reaction with the information given

$$ \Delta H^\circ - T\Delta S^\circ = {-RT\ln{K_p}} $$

$$ \Delta S ^\circ = {-RT\ln{K_p} - \Delta H ^\circ \over T} $$

$K_p = 1.3*10^3$ is given and $\Delta H^\circ = -66.5 kJ/mol$ since that is the change in enthalpy when 1 mole of the dimer is formed from each molecule of acetic acid. Inputting these into the equation equals:

$$ \Delta S^\circ = {-(8.314)(298)\ln{(1.3*10^3)} - (-6.65*10^4) \over (298)} $$

$$ \Delta S^\circ = {163.5 J/mol} $$

This reaction is both entropically and enthaplically favorable so it is a spontaneous reaction.

	\(\ce{[HNO2]}\)	\(\ce{[H3O+]}\)	\(\ce{[NO2^-]}\)
Initial	\(0.75 M\)	\(0\)	\(0.030M\)
Change	\(-x\)	\(+x\)	\(+x\)
Equilibrium	\(0.75-x\)	\(x\)	\(0.03+x\)