The equilibrium constant under standard conditions (1 atm, 298 K, pH 7.0) for th

Question

The equilibrium constant under standard conditions (1 atm, 298 K, pH 7.0) for the reaction below, catalyzed by fumarase, is 4.00. At 310 K, the equilibrium constant is 8.00.

Fumarate(aq) + H2O 1-malate (aq)

a. What is the standard Gibbs energy change G0 for this reaction?

b. If [1-malate] = 0.01M and [fumarate] = 0.02 M, what is the Gibbs energy change at 298 K?

c. From b. above, how will this reaction proceed to equilibrium – that is: will more fumarate or Lmalate form?

d. Evaluate H0 and S0 for this reaction at 298K. State any assumptions you make.

Explanation / Answer

a )  Go = - RT lnK

= - 8.314 x 10^-3 x 298 x ln 4

= - 3.43 kJ / mol

b)

K = [1-malate] / Fumarate

K = 0.01 / 0.02

K = 0.5

RT lnK =  8.314 x 10^-3 x 298 x ln 0.5

= -1.717

G = Go + R T ln K

= -3.43 + (-1.717)

= -5.147 kJ/mol

c)

G value is negative so equilibrium shift to forward direction . more malate will form

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