Question
Please answer the following question and show all necessary work to receive credit:
where Delta G degree ' = the standard free energy charge for the reaction when all reactants and products are in their standard states. The standard state is defined as a concentration of 1 M at pH 7. [H] is defined to be 1 at pH 7. [H2O] is also defined to be 1, which means that it may be omitted from the expression for the ratio. R = 8.315 times 10-3 kJ/mol-K T = 298 K [ ] means concentration in mol/L or molarity (M) For the reaction When all reactants and products are in their standard states, Q= 1, In 1 = 0, and Delta G = Delta G degree '. At equilibrium. AG = 0 and Q = K' eq (the equilibrium constant for the reaction) which leads to the equation The equilibrium ratio of [Products]/[Reactants] is the maximum value of the ratio for which the reaction is exergonic delta G O) or can occur. Homework assignment should verify this. Endergonic under standard conditions of equilibrium favors reactants (Keq 1) Use the following data to calculate the equilibrium constant (Equation 3) and delta G (Equation J) at the six given [Products]/[Reactants]] ratios. Show your work and put your answers for Keq and delta G in the table. Delta G=+B.Okj/mol R=8.315x103kj/mol.K T=298 K RT = State your conclusions from the above data by underlining one of the two choices in each parentheses. Delta G is negative for all values of [Products]/[Reactants] which are (less than or greater than) K'eq, meaning that these are the conditions under which the reaction (can or cannot) occur. An endergonic reaction can be changed to an exergonic reaction by increasing the (reactants or products) and/or by decreasing the (reactants or products)
Explanation / Answer
d will be used for delta
Equilibrium constant K is same at same temperature
dG0 = -RTlnK
(5.0) = -(0.008315)(298)lnK
lnK = -2.01786
K = 0.1329393705
In first column,
K/2 = 0.0664696853
K/3 = 0.0443131235
I will answer remainder by rows
Row 1
G = G0 + RTlnK
= 5.0 + (0.008315)(298)ln(3)
= 7.72 kJ/mol
Endergonic
Row 2
G = G0 + RTlnK
= 5.0 + (0.008315)(298)ln(2)
=6.72 kJ/mol
Endergonic
Row 3
G = G0 + RTlnK
= 5.0 + (0.008315)(298)ln(1)
= 5.0 kJ/mol
Endergonic
Row 4
G = G0 + RTlnK
= 5.0 + (0.008315)(298)ln(K)
=0 kJ/mol
Equilibrium
Row 5
G = G0 + RTlnK
= 5.0 + (0.008315)(298)ln(K/2)
=-1.72 kJ/mol
Exergonic
Row 6
G = G0 + RTlnK
= 5.0 + (0.008315)(298)ln(K/3)
=-2.72 kJ/mol
Exergonic
Statement question
less than, can
reactant, product
Comment if there is any ambiguity or further question