Question

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Consider the reaction of silver ion with bromide ion given below at 25°C. Ag+ (aq) + Br+ (aq) → AgBr (s) What is the value of AGº for this reaction, if the equilibrium constant, Keq = 2.0 x 1012? A) -5.9 kJ/mol B) -70 kJ/mol C) -5.9 x 103 kJ/mol D) -7.0 104 kJ/mol E) -0.69 kJ/mol The rest mass of a proton is 1.00727252 u and that of a neutron is 1.008665 u. The He nucleus weighs 4.00150286 u. Calculate the total binding energy of the nucleus. A) 1.667 x 10-7 joule B) 2.266 × 10-12 joule C) 3.861 10-11 joule D) 4.532 x 10-12 joule E) 4.532 x 10-10 joule

Answer 1

The standard Gibbs free energy (AG) and equilibrium constant (keq) for a particular temperature (T) are related, as follows: AG= RT In Kea ( R = 8,314 3/k/mal) here, Given that , Kon=2x10 a 1 314 x 29 "o ket = 2x 10² at 25°c is T= 25°C = (25 +273) K = 298K Now, Aci= - RT ln keq os, dh'= - 8.314 3. It! mal' x 298 k ln (2x102) an, age = -f3:314 x298* An (2x16")} J. mati! Or, AG = - 70.188103 Ilmal (Since, 1 kJ = 103) on, AG = -70.18 kJ/mol. - 70 kJ/mal. . . Corred option : B)-70 kJ/mal.) # Binding energy (E) and mass defect (Am) are related as fallows: here, E = o a velocity of light = 3x10 m/sec no. of Proton = 2 no al neutron = (4-2) = 2 . Calculated mass = ax (mass af neutron) + 2x (mass af proton) ={2* 1.008665 +241.007272 52} U = 4.031875040 8. Mass defect (Am) = Calculated mass actual mass = (4.03187504 -4.00150286) u = 0.030372180 (10= 1.66054*1727 kg) -0.0303721871660548 1627 kg E = Amě = 0.03037218*1166054x1027 kq x(3x109) 75 = 4.532*10'? J (Since, 13 = 1kg.m./6") :. Correct option i 16.) 41532 4161 J um am "He