following reaction: glycine + glycine glycylglycine + HO Use the data in 4G, and
ID: 1053038 • Letter: F
Question
following reaction: glycine + glycine glycylglycine + HO Use the data in 4G, and the equilibrium constant at 298 K, keeping in mind that the reaction is carried out in an aqueous buffer solution. Assume that the value of A,G is essen- tially the same at 310 K. What conclusion can you draw about your result? 633 From the following reactions at 25°C fumarate2 NH -aspartate G" =-36.7 kJ mol-1 fumarate2-+ H2O malate?- A,G", =-2.9 kJ mol-1 calculate the standard Gibbs energy change and the uiibium constant for the following reaction: equilibrium malate?-+NHJ-aspartate" + H2O dom Pol dom coil forms. 6.34 A 34 A polypeptide can exist in either the helical or ran- l form librium s. The equilibrium constant for the equi- lbrium eT tion values of reaction of the helix to the random coil transi- n is 0.86 a u6 at 40°C and 0.35 at 60°C. Calculate the ,Ho and ,S° for the reaction.Explanation / Answer
to solve 6.32 , data from appendix 2 is needed.
#6.33
The given reactions are
i. Fumarate^2- + NH4+ -----------> Aspartate^- delta G = -36.7 kJ/ mol
ii. Fumarate^2- + H2O -------------> Malate ^2- delta G = -2.9 kJ/ mol
We want to find deltaG for the reaction, Malate ^2- + NH4+ ------> Aspartate^- + H2O
We will use Hess's law to solve this.
In the required reaction, Malate^2- is on reactant side but in the given reaction #ii, it's on product side. So we will flip equation (ii) to bring Malate^2- on reactant side.
iii . Malate ^2- -----------------> Fumarate^2- + H2O delta G = 2.9 kJ/ mol ( sign of delta G is also reversed)
Let's add equation i & iii
Fumarate^2- + NH4+ -----------> Aspartate^- delta G = -36.7 kJ/ mol
Malate ^2- -----------------> Fumarate^2- + H2O delta G = 2.9 kJ/ mol
---------------------------------------------------------------------------------------------------------------------
Malate^2- + NH4+ ---------> Aspartate^- + H2O delta G = -33.8 kJ / mol ( Fumarate^2- is cancelled because it is present on both sides)
Standard Gibb's energy change delta Go = -33.8 kJ/ mol
delta Go and equilibrium constant K are related by the following equation
Delta Go = - RT ln K
We will convert delta G from kJ to J
therefore we have -33.8 kJ * 1000 J/ 1 kJ = -33800 J
Following data is known to us
delta Go = -33800 J
R = 8.314 J/ mol K
T = 25 C = 25 + 273 K = 298 K
Substituting these values , we get
-33800 J = - 8.314 J / mol K * 298 K * ln K
ln K = -33800 J / - 8.314 J/mol K * 298 K
ln K = 13.64
K = e^13.64
K = 8.41 x 10^5
Equilibrium constant for the given reaction is 8.41 x 10^5
#6.34
The equilibrium reaction can be written as
Helical from <-------------> Random coil form
Following data is given to us
K1 = 0.86
T1 = 40 C = 313 K
K2 = 0.35
T2 = 60 C = 333 K
These variables are related to each other by Vant Hoff equation which is given as
ln ( K2/K1) = Delta Ho/R * ( 1/T1 - 1/T2)
ln ( 0.35/0.86) = delta Ho/ 8.314 J/ mol K * ( 1/313K - 1/333K) Here, K gets cancelled
ln 0.40698 = delta Ho/ 8.314 J/mol * ( 0.000191885)
-0.89899 = delta Ho / 8.314 J/mol * 0.000191885
delta Ho = -0.89899/ 0.000191885 * 8.314 J/ mol
delta Ho = -3.895 x 10^4 J/mol
on rounding to 2 sig figs we get
Delta Ho = -3.9 x 10^4 J/ mol
Delta Ho & Delta So are related to each other by following equation
Delta Go = Delta Ho - T Delta So
In order to solve for Delta So we first need to find Delta G using one of the given K and T
Delta Go = - RT ln K
Let's use K1 and T1 in above equation
we get, Delta Go = - 8.314 J/ mol K * 313 K * ln 0.86
Delta Go = 392.48 J
Substituting this in equation for delta So, we get
Delta Go = Delta Ho - T Delta So
392.48 J/mol = - 3.9 x 10^4 J/mol - 313K* deltaSo ( we will use same value of T that we used for calculating delta G )
delta So = 39392.48 J/mol /- 313 K = - 126 J/mol K
On rounding to 2 sig figs, we get
Delta So = -130 J/K
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