10. What is the value of delta G when [H3O+]= 6.7 x 10^-9M, [CH3NH3]= 2.4 x 10^-
ID: 1058892 • Letter: 1
Question
10. What is the value of delta G when [H3O+]= 6.7 x 10^-9M, [CH3NH3]= 2.4 x 10^-3M, and [CH3NH2] = 0.098M?11. Calculate delta H° for the following reaction: BaCO3(s) ---> BaO(s) + CO2(g)
12. Calculate delta S° for the following reaction: BaCO3(s) ----> BaO(s) + CO2(g)
13. Calculate delta G° for the following reaction: 10. The K, for methylamine (CH3NH2) at 25 oC is 4.4 x 10 What the value of AG when 1-6.7 x x is [H,o 10 M [CH, NH, J-2.4 x 103 M, and [CH3NH2]- 0.098 M? 11. Calculate AH for the following reaction: Baco, (s) Bao(s) CO2 (g) 12. Calculate AS for the following reaction: 13. Calculate AG for the following reaction: BaCO3 (s) Bao(s) Co20g)
Explanation / Answer
10. Write down the reaction as
CH3NH3+ (aq) + H2O (l) -------> CH3NH2 (aq) + H3O+ (aq)
The acid dissociation constant Ka is given as
Ka = [CH3NH2][H3O+]/[CH3NH3+]
Given Kb, we can calculate Ka as Ka*Kb = Kw
===> Ka*(4.4*10-4) = 1.0*10-14
===> Ka = (1.0*10-14/4.4*10-4) = 2.273*10-11
Given Ka, we can find G0 as
G0 = -R*T*ln Ka = -(8.314 J/mol.K)*(298 K)*ln (2.273*10-11) = 60718.688 J/mol = 60.718 kJ/mol 60.72 kJ/mol.
The reaction quotient Q for the above reaction is
Q = [CH3NH2][H3O+]/[CH3NH3+] = (0.098)*(6.7*10-9)/(2.4*10-3) = 2.736*10-7
The change in free energy for the above concentrations is given as
G = G0 + RTlnQ = (60.72 kJ/mol) + (8.314 J/mol.K)*(298 K).ln(2.736*10-7) = (60.72 kJ/mol) + (-37440.074 J/mol) = (60.72 kJ/mol) – (37.44 kJ/mol) = 23.28 kJ/mol (ans).
11. I shall use values obtained from internet for this portion:
Hf0 (BaCO3) = -1218.8 kJ/mol
Hf0 (BaO) = -558.1 kJ/mol
Hf0 (CO2) = -393.5 kJ/mol
Enthalpy of reaction, Hrxn0 = Hf0 (products) – Hf0 (reactants) = [(-558.1) + (-393.5)] kJ/mol – (-1218.8 kJ/mol) = -951.6 kJ/mol + 1218.8 kJ/mol = 267.2 kJ/mol (ans).
12. Again, I shall use standard entropy of formation values from internet:
Sf0 (BaCO3) = 112.13 J/mol.K
Sf0 (BaO) = 72.09 J/mol.K
Sf0 (CO2) = 213.8 J/mol.K
Entropy of formation, Srxn0 = Sf0 (products) - Sf0 (reactants) = [(72.09) + (213.8)] J/mol.K – (112.3 J/mol.K) = 173.59 J/mol.K (ans).
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