For carbonic acid (H_2CO_3), K_a1 = 4.30 Times 10^-7 and K_a2 = 5.62 Times10^-11
ID: 973507 • Letter: F
Question
For carbonic acid (H_2CO_3), K_a1 = 4.30 Times 10^-7 and K_a2 = 5.62 Times10^-11 at 298 K. calculate the pH at 298 K for a 0.24 M solution of disodium carbonate (Na_2CO_3). Consider the reaction: 2 NO_2 (g) N_2O_4(g) The standard state Gibbs free energies of formation for NO_2 (g) and N_2O_4 (g) are 51.84 and 98.00 kJ mol^-1, respectively. What is Delta G degree for this reaction? Calculate the equilibrium constant for the reaction at 25 degree C. What is Delta G for the reaction at 25 degree C when both gases are at 1.5 atm pressure? As illustrated below, a strip of copper is placed m a 1.20 M solution of copper (II) nitrate and a strip of silver is placed in a 1.20 Times 10^-3 M solution of silver (I) nitrate. The two metal strips are connected to a voltmeter by wires and a salt bridge connects the solutions. Assuming that the electron flow is spontaneous from left to right as shown, what is the net cell reaction? What is the value of epsilon_cell? What is the value of epsilon_cell at the stated concentrations and 298 K ? What is the value of Delta G for the overall cell reaction at 298 K ? What maximum amount of work can the reaction yield?Explanation / Answer
Ans for 6)
Given transformation,
2NO2 (g) <--------> N2O4 (g)
Give thermodynamic data:
0G (NO2(g)) = +51.84 kJ/mol
0G (N2O4 (g)) = +98.00 kJ/mol
a) 0GRXN =?
Formula,
0GRXN = 0G of products - 0G of reactants.
0GRXN = [0G (N2O4 (g))] – [2 x 0G (NO2(g)]
0GRXN = (+98.00)- (2x +51.84)
0GRXN = (+98.00) –(103.68)
0GRXN = - 5.68 kJ/mol
0GRXN = - 5680 J/mol.
=======================
b) Relation between equilibrium constant Keq and 0GRXN is,
0GRXN = -RT ln(Keq)
We have T= 25 0C = 298.15 K, R = 8.314 J/K.mol, 0GRXN = - 5680 J/mol.
Let us put in above equation and solve it for Keq.
-5680 = - 8.314 x 298.15 x ln(Keq)
5680 = 2478.8 ln(Keq)
ln(Keq) = 5680/2478.8
ln(Keq) = 2.291
Keq = e2.291
Keq = 9.89
Equilibrium constant for given reaction at 25 0C is 9.89.
================================
c) Given transformation,
2NO2 (g) <--------> N2O4 (g)
n = Number of gaseous products – Number of gaseous reactant
n = 1 – 2
n = -1
Kp = PN2O4/(PNO2)2
Kp = 1.5/(1.5)2 ………… (given data)
K= 1/1.5
K = 1/(3/2)
K = 2/3
Kp = 0.67
Again we have,
0GRXN = -RT ln(Kp)
And, Kp = 2/3 =0.67, T=298.15 K and R = 8.314 J/K.mol.
0GRXN = -8.314 x 298.15 x ln(2/3)
0GRXN = -8.314 x 298.15 x (-0.4055)
0GRXN = 1005.1 J/mol
0GRXN = 1.005 kJ/mol.
Gibbs free energy change at given condition is, +1.005 kJ/mol.
==========================XXXXXXXXXXXXXXXXXXX==============================
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.