Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.

Lesson: Equilibrium Constants and Gibbs Free Energies

Worksheet • 8 Questions

Q1:

Carbon tetrachloride is formed by the chlorination of methane at high temperature: The standard entropies and enthalpies of formation for the reactants and products are shown in the table.

Material Standard Molar Entropy ๐‘† โฆต ๏Šจ ๏Šฏ ๏Šฎ (J/Kโ‹…mol) Standard Enthalpy of Formation ฮ” ๐ป โฆต f (kJ/mol)
C C l ( ) 4 l 214.4 โˆ’ 1 2 8 . 2
C C l ( ) 4 g 309.7 โˆ’ 9 5 . 7
C H ( ) 4 g 186.3 โˆ’ 7 4 . 6
C l ( ) 2 g 223.1 0.0
H C l ( ) g 186.9 โˆ’ 9 2 . 3

Assuming the thermodynamic parameters do not vary with temperature, calculate, to 1 significant figure, the equilibrium constant for this reaction at 2 5 0 โˆ˜ C .

  • A 7 ร— 1 0 ๏Šฉ ๏Šญ
  • B 9 ร— 1 0 ๏Šฉ ๏Šฎ
  • C 1 ร— 1 0 ๏Šฉ ๏Šฌ
  • D 7 ร— 1 0 ๏Šฉ ๏Šฌ
  • E 1 ร— 1 0 ๏Šช ๏Šฆ

Q2:

Which of the following is the best definition of ฮ” ๐บ โฆต , the standard change in Gibbs free energy for a reversible process?

  • AWhen the concentrations or partial pressures of reactants and products are all equal to the standard value, ฮ” ๐บ โฆต is the energy that must be absorbed for equilibrium to be reached.
  • B ฮ” ๐บ โฆต is the difference in free energy between the pure products in their stoichiometric ratio and the equilibrium mixture of reactants and products.
  • CWhen the concentrations or partial pressures of reactants and products are all equal to the standard values, ฮ” ๐บ โฆต is the energy that must be released for equilibrium to be reached.
  • D ฮ” ๐บ โฆต is the difference in free energy between the pure reactants in their stoichiometric ratio and the equilibrium mixture of reactants and products.
  • EWhen the sum of the concentrations or partial pressures of reactants and products is equal to the standard value, ฮ” ๐บ โฆต is the energy that must be released for equilibrium to be reached.

Q3:

Consider the reaction g l y c e r o l ( ) + H P O ( ) D L - g l y c e r o l - - p h o s p h a t e + H O ( ) a q a q l 4 2 โ€“ 2 1 ( ๐‘Ž ๐‘ž ) 2 โˆ’ . The equilibrium constant for this reaction at 3 7 โˆ˜ C is 0.0120, and the value of ฮ” ๐บ โฆต r at 2 5 โˆ˜ C is 9.37 kJ. Given this information, what is the equilibrium constant ( ๐พ ) e q for the reaction at 2 5 โˆ˜ C , and what is the value of ฮ” ๐ป โฆต r for the reaction? (You may assume that the ฮ” ๐ป โฆต r of the reaction is independent of temperature.)

  • A ๐พ = 0 . 0 2 3 e q at 2 5 โˆ˜ C ; ฮ” ๐ป = โˆ’ 4 1 . 2 โฆต r k J
  • B ๐พ = 0 . 4 6 2 e q at 2 5 โˆ˜ C ; ฮ” ๐ป = 4 1 . 2 โฆต r k J
  • C ๐พ = 0 . 0 0 3 e q at 2 5 โˆ˜ C ; ฮ” ๐ป = โˆ’ 2 1 . 2 โฆต r k J
  • D ๐พ = 0 . 0 4 6 e q at 2 5 โˆ˜ C ; ฮ” ๐ป = 2 1 . 2 โฆต r k J

Q4:

Hydrogen sulfide reacts reversibly with sulfur dioxide to form sulfur and water: The standard free energies of formation, ฮ” ๐บ โฆต ๏Œฟ , for hydrogen sulfide and other materials are shown in the table. These energies are measured at 2 5 โˆ˜ C .

Material H S ( ) 2 g S O ( ) 2 g S O ( ) 3 g S ( ) 8 s S ( ) g H O ( ) 2 g H O ( ) 2 l
Standard Gibbs Free Energy of Formation ฮ” ๐บ โฆต ๏Œฟ (kJ/mol) โˆ’ 3 3 . 4 โˆ’ 3 0 0 . 1 โˆ’ 3 7 1 . 1 0.0 238.3 โˆ’ 2 2 8 . 6 โˆ’ 2 3 7 . 1

Calculate, to 2 significant figures, the equilibrium constant for this reaction at 2 5 โˆ˜ C .

  • A 6 . 3 ร— 1 0 ๏Šง ๏Šฎ
  • B 4 . 5 ร— 1 0 ๏Šจ ๏Šช
  • C 2 . 3 ร— 1 0 ๏Šฌ
  • D 1 . 4 ร— 1 0 ๏Šฉ
  • E 6 . 6 ร— 1 0 ๏Šง ๏Šซ

A student uses the values of ฮ” ๐บ โฆต ๏Œฟ at 2 5 โˆ˜ C to calculate the equilibrium constant for this reaction at 9 0 โˆ˜ C . Why are the results of this calculation likely to be inaccurate?

  • A ฮ” ๐บ โฆต ๏Œฟ values change with temperature.
  • BThe products of the reaction have different phases at the higher temperature.
  • CThe standard pressure changes with temperature.
  • DThe reaction may not occur at the higher temperature.
  • EThe equation ฮ” ๐บ = โˆ’ ๐‘… ๐‘‡ โฆต l n ๐พ is only true at a standard temperature of 2 5 โˆ˜ C .

Q5:

The standard change in Gibbs free energy for a reversible process, ฮ” ๐บ โฆต , is measured for a standard solution of reactants and products. What are the initial concentrations of the reactants and products in this solution?

  • AThe concentration of each reactant and product is 1 M.
  • BReactants and products are present in the stoichiometric ratio and the sum of the reactant concentrations is 1 M.
  • CReactants are present in the stoichiometric ratio and the sum of the reactant concentrations is 1 M. Product concentrations are defined in the same way.
  • DReactants and products are present in the equilibrium ratio and the sum of the reactant and product concentrations is 1 M.
  • EReactants and products are present in the stoichiometric ratio and the sum of the reactant and product concentrations is 1 M.

Q6:

The equilibrium constant ( ๐พ ) ๐‘ for the hydrolysis of adenosine triphosphate (ATP) to adenosine diphosphate (ADP) and phosphate is 1 . 6 7 ร— 1 0 / 5 3 m o l d m at 3 7 โˆ˜ C , and the ฮ” ๐ป ๐‘Ÿ โฆต for this reaction has a value of โˆ’ 2 0 . 1 kJ/mol. Given this information, what is the value of ฮ” ๐‘† ๐‘Ÿ โฆต for the reaction?

Q7:

Consider the following two reactions: (1) g l u t a m a t e + N H g l u t a m i n e 4 + , for which ฮ” ๐บ = 1 5 . 7 / โฆต ๐‘Ÿ k J m o l at 3 7 โˆ˜ C ; and (2) A T P A D P + P i , for which ฮ” ๐บ = โˆ’ 3 1 . 0 / โฆต ๐‘Ÿ k J m o l at 3 7 โˆ˜ C . (In reaction (2), A T P denotes adenosine triphosphate, A D P denotes adenosine diphosphate, and P i represents inorganic phosphate.) These two reactions can be coupled by an enzyme catalyst (glutamine synthetase), which leads to the following reaction: (3) g l u t a m a t e + N H + A T P g l u t a m i n e + A D P + P i 4 + . What is the equilibrium constant for reaction (3) at a temperature of 3 7 โˆ˜ C ?

  • A 3 . 8 ร— 1 0 2
  • B 2 . 3 ร— 1 0 โˆ’ 3
  • CThere is not enough information provided to answer this question.
  • D1.1

Q8:

Under certain conditions, gaseous ammonia can decompose into nitrogen and hydrogen gases: The partial pressures of N H 3 , H 2 , and N 2 are denoted ๐‘ƒ ( N H ) 3 , ๐‘ƒ ( H ) 2 , and ๐‘ƒ ( N ) 2 respectively. When the initial partial pressure of each gas is equal to the standard value of 1.00 atm and the temperature is fixed at 298 K, the change in Gibbs free energy for the reaction, ฮ” ๐บ โฆต , is 33.00 kJ/mol. Calculate, to 3 significant figures, the change in Gibbs free energy, ฮ” ๐บ , at 298 K when ๐‘ƒ = 1 2 . 9 ( N H ) a t m 3 , ๐‘ƒ = 0 . 2 5 0 ( H ) a t m 2 , and ๐‘ƒ = 0 . 8 7 0 ( N ) a t m 2 .

Preview