Consider the reaction X + Y Z. Based solely upon this information, do you expect

Question

Consider the reaction X + Y Z. Based solely upon this information, do you expect the entropy change in this reaction to be less than, equal to, or greater than 0?  

A

less than 0..

B

equal to 0.

C

greater than 0.

Consider the phosphoglucoisomerase reaction: Glucose-6-P = Fructose-6-P          G'0 = +1,790 J/mol

At equilibrium, is the concentration of glucose-6-P greater than or equal to or less than the concentration of fructose-6-P?  

If the concentration of fructose-6-P in a cell is greater than the concentration of glucose-6-P, is the value of G' greater than or equal to or less than 0 J/mol.

A

less than 0..

B

equal to 0.

C

greater than 0.

Explanation / Answer

Ans. 1. The number of moles of reactants = 1 mol (X) + 1 mol (Y) = 2 mol

The number of moles of product = 1 mol (Z)

Note that the number of moles of product is lesser than the number of moles of reactants. Therefore, the entropy of the system decreases because 1 mol product has lesser degree of freedom than 2 mol reactants.

Therefore, entropy decreases and is less than 0.

Correct option. A. Less than 0.

Ans. 2. Using the equation dG = dG0’ + RT lnK            - equation 1

            Where, dG = calculated/ experimental free energy change = ?

                                    dG0’ = standard/ theoretical free energy change                                                            R = 0.0083146 kJ mol-1K-1

                                    T = temperature in kelvin = (0C + 273.15) K

                                    K = equilibrium constant under given condition

Putting the values in equation 1-

            1.790 kJ/ mol = - (0.0083146 kJ mol-1K-1) x 298 K x ln K

            Or, 1.790 kJ/mol / (-2.4777508 kJ/mol) = ln K

            Or, -0.72243 = 2.303 log K

            Or, log K = -0.72243 / 2.303 = -0.31369

            Or, K = antilog (-0.31369) = 0.486

Therefore, equilibrium constant, K = 0.486

Now, from the balanced reaction, equilibrium constant, K = [F-6-P] / [G-6-P]

Or, 0.486 = [F-6-P] / [G-6-P]

            Or, [F-6-P] = 0.486 x [G-6-P]

Therefore, [F-6-P] < [G-6-P]         - at equilibrium.

Or, [G-6-P] is greater than that of [F-6-P].

Ans. 3. If [F-6-P] > [G-6-P], then equilibrium constant K is always greater than 1.

Say, [F-6-P]/ [G-6-P] = 2, that is [F-6-P] is double that of [G-6-P].

So, K = [F-6-P]/ [G-6-P] = 2

Putting the values of K=2 in equation 2-

                        dG = dG0’ + RT lnK             - equation 2

            Where, dG = calculated/ experimental free energy change = ?

                                    dG0’ = standard/ theoretical free energy change                                                            R = 0.0083146 kJ mol-1K-1

                                    T = temperature in kelvin = (0C + 273.15) K

                                    K = equilibrium constant under given condition.

            dG = 1.790 kJ/mol + (0.0083146 kJ mol-1 K-1) x 298 K x (ln 2)

            or, dG = 1.790 kJ/mol + 1.72 kJ/mol = 3.51 kJ/mol

Therefore, dG > dG0            – when [F-6-P] > [G-6-P]

And, dG is greater than 0.

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