I am working on question number 10 is 33 kJ mol-. Calculate the boiling point of
ID: 556306 • Letter: I
Question
I am working on question number 10
is 33 kJ mol-. Calculate the boiling point of toluene at an atmospheric pressure of 0.90 bar. (9) The acid dissociation constant of phosphoric acid is 7.5 × 10-3 at 25 C. What is the standard free energy of formation of the aqueous H;PO, molecule? (10) Ribonuclease has two conformations, only one of which is active. At 50 °C, the equilibrium concentrations of the active and inactive forms are, respectively, 9.97 × 10-4 and 2.57 × 10-6 mol L-1 in a solution with a total molarity of 1 mmol L-1. At 100 °C, the equilibrium concentrations of the active and inactive forms are 8.6 × 10 and 1.4 × 10-4 mol L-1 (a) What is rHo for the process that converts the active form to its inactive confor- -4 mation? (b) What is the ratio of the active to the inactive form at 37 C? k (HH) is a naturally occurring mutant form of hemoglobin. It ally exists in solution as a dimer (two copies of the protein associated together) 11) Hemoglobin Howic but can also form a tetramer:Explanation / Answer
(9) We know the relation between the Gibbs free energy and equilibrium constant. The below equation is valid only in equilibrium condition.
G = -RTlnKeq ………….(1)
G = Gibbs free energy
R = Gas constant = 8.314 J/mol-K
T = Temperature =25o C = (273+25) K = 298 K (Given)
Keq = equilibrium constant
Equation for the Dissociation of phosphoric acid
H3PO4 = H+ + H2PO4- {Equilibrium Equation}
Ka = 7.5x10-3 (Given)
Formation of the aqueous H3PO4 is
H+ + H2PO4- = H3PO4
Kf = 1/Ka = 1/7.5x10-3
Putting all the values in equation (1), we get the standard free energy of formation of aqueous H3PO4 molecule.
Gof = -8.314x298xln(1/7.5x10-3)
= -12122.39 J/mol
= -12.122 kJ/mol
Hence the standard free energy formation of the aqueous H3PO4 is -12.122 kJ/mol.
(10) (a) Equilibrium equation for the conversion of the active form of Ribonuclease to its inactive conformation.
Ribonuclease (active) = Ribonuclease (inactive)
Keq = [Ribonuclease (inactive)]/[ Ribonuclease (active)] …..(1)
To solve this question, we have to use the following equation,
ln[(Keq)2/(Keq)1]= -H/R[(1/T2) – (1/T1)] (van’t Hoff equation)….(2)
Where, H = enthalpy of reaction
Given, T1 = 50o C = (50+273) K = 323 K
[Ribonuclease (inactive)] = 2.57x10-6 molL-1
[Ribonuclease (active)] = 9.97x10-4 molL-1
Using equation (1) (Keq)1 = 2.57x10-4 molL-1/9.97x10-6 molL-1
= 2.58x10-3
Similarly, T2 = 100o C = (100+273) K = 373 K
[Ribonuclease (inactive)] = 1.4x10-4 molL-1
[Ribonuclease (active)] = 8.6x10-4 molL-1
(Keq)2 = 1.4x10-4 molL-1/8.6x10-4 molL-1
= 1.63x10-1
Know, Using van’t Hoff equation, we get
ln[1.63x10-1/2.58x10-3]= - rHo/8.314[(1/373)-(1/323)]
4.146 = - rHo/8.314[-50/(373x323)]
rHo = 4.146x373x323x8.314/50
= 83057.85 J/mol
= 83.058 kJ/mol
Hence, rHo is 83.058 kJ/mol for the process that converts the active form to its inactive conformation.
(b) Using the van Hoff’s equation again, we get the ratio of the active to inactive form at 37o C.
Let’s Suppose, T2 = 37o C = (273+37) K = 310 K
(Keq)2 = ?
And T1 and (Keq)1 are same as used in part (a),
rHo = 83.058 kJ/mol = 83058J/mol
Putting in equation (2),
ln[(Keq)2/2.58x10-3]= -83058/8.314[(1/310)-(1/323)]
(Keq)2 = 2.58x10-3 x exp(-1.297)
(Keq)2 = 2.58x10-3 x 10(-1.297/2.303)
(Keq)2 = 2.58x10-3 x 10-0.563
(Keq)2 = 2.58x10-3 x 0.274 = 0.7057x10-3 = 7.057x10-4
We know that (Keq)2 is the ratio of inactive conformer to active conformer. (see equation (1))
Then,
[Ribonuclease (active)]/[ Ribonuclease (inactive)]=1/(Keq)2=1/(7.057x10-4)
= 1.42x103
Hence, the ratio of the active to the inactive form at 37o C is 1.42x103.
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