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Given that the Δ G f ° for Pb 2+ ( aq ) and Cl ( aq ) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, K sp , for PbCl 2 ( s ).

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Determine the standard free energy change, Δ G f ° , for the formation of S 2− ( aq ) given that the Δ G f ° for Ag + ( aq ) and Ag 2 S( s ) are 77.1 k/mole and −39.5 kJ/mole respectively, and the solubility product for Ag 2 S( s ) is 8 × 10 −51 .

90 kJ/mol

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Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.

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The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
H 2 O ( l ) H 2 O ( g ) Δ G 298 ° = 8.58 kJ

(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?

(b) Determine the equilibrium constant, K P , for this physical process.

(c) By calculating ∆ G , determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, P H 2 O , is 0.011 atm.

(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of P H 2 O in the air?

(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b) K p = 0.031; (c) The evaporation of water is spontaneous; (d) P H 2 O must always be less than K p or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.

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In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:
Glu + ATP G6P + ADP Δ G 298 ° = −17 kJ

In this process, ATP becomes ADP summarized by the following equation:
ATP ADP Δ G 298 ° = −30 kJ

Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
Glu G6P Δ G 298 ° = ?

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One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
G6P F6P Δ G 298 ° = 1.7 kJ

(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?

(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M , however, in a typical cell, they are not even close to these values. Calculate Δ G when the concentrations of G6P and F6P are 120 μ M and 28 μ M respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.

(a) Nonspontaneous as Δ G 298 ° > 0 ; (b) Δ G 298 ° = R T ln K , Δ G = 1.7 × 10 3 + ( 8.314 × 335 × ln 28 128 ) = −2.5 kJ . The forward reaction to produce F6P is spontaneous under these conditions.

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Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.

(a) N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g )

(b) HCl ( g ) + NH 3 ( g ) NH 4 Cl ( s )

(c) ( NH 4 ) 2 Cr 2 O 7 ( s ) Cr 2 O 3 ( s ) + 4H 2 O ( g ) + N 2 ( g )

(d) 2Fe ( s ) + 3O 2 ( g ) Fe 2 O 3 ( s )

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When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of Δ G , Δ H , and Δ S for this process, and justify your choices.

Δ G is negative as the process is spontaneous. Δ H is positive as with the solution becoming cold, the dissolving must be endothermic. Δ S must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.

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An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu 2 S decomposes to form copper and sulfur described by the following equation:
Cu 2 S ( s ) Cu ( s ) + S ( s )

(a) Determine Δ G 298 ° for the decomposition of Cu 2 S( s ).

(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine Δ G 298 ° for the process.

(c) The production of copper from chalcocite is performed by roasting the Cu 2 S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.

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What happens to Δ G 298 ° (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?

(a) S ( s ) + O 2 ( g ) SO 2 ( g )

(b) 2SO 2 ( g ) + O 2 ( g ) SO 3 ( g )

(c) HgO ( s ) Hg ( l ) + O 2 ( g )

(a) Increasing P O 2 will shift the equilibrium toward the products, which increases the value of K . Δ G 298 ° therefore becomes more negative.
(b) Increasing P O 2 will shift the equilibrium toward the products, which increases the value of K . Δ G 298 ° therefore becomes more negative.
(c) Increasing P O 2 will shift the equilibrium the reactants, which decreases the value of K . Δ G 298 ° therefore becomes more positive.

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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