Question

A 0.0403 M solution of a particular monoprotic weak acid. HA, has a pH of 3, 40 a 298 K. What is Delta G^Compositefunction for the following equilibrium? HA (aq) + H_2O(l) rightarrow leftarrow H_2O(aq) = A(aq) A) 7.95 kJ B) 30.8 kJ C) 11.4 kJ D) 19.4 kJ For a reversible phase change at constant temperature and pressure. A) Delta U = 0. B) Delta H = 0. C) w = 0 D) q = 0 E) Delta G = 0 For a certain process, at 300. K, Delta G -22.0 kJ and Delta H = -7.0 kJ. If the process is carried out irreversibly, what is the amount of useful work that can be performed? A -29.0 kJ B) -7.0 kJ C) -15.0 kJ D) 22.0 kJ The third law of thermodynamics states that A) the entropy of the universe is increasing. B) the entropy of the universe is constant. C) the entropy is zero at 0 K for a perfect crystal. D) the absolute entropy of a substance decrease with increasing the temperature. E) the entropy of the universe equals the sum of the entropy system and that of the surroundings. What is the change in entropy when 0.374 g of silicon is burned in excess oxygen to yield silicon dioxide at 298 K? Si(s) + O_2(g0 rightarrow Delta S^Compositefunction = - 182.4 J/K at 298 K A) -68.2 J/K B) -1.37 times 10^3 J/K C) -2.43 J/K D)-182 J/K E) -8.15 times 10^-3 J/K

Answer 1

"Given equilibrium HA_(aq) + H_2 O doubleheadedarrow H_3 O^+ + A^- _(aq) given data: [HA] = 0.0403 M P^H = 3.40: T = 298 we need to find out delta G degree = ? since p^H = 3.40: we can calculate [H^+] by using the equation [H^+] = 10^- pH [H^+] = 10^- 3.40 = 0.000398 and now we need to calculate k_a = ? According to the general definition of k_a: k_a = [A^-][H_3 O^+]/[HA] In this casee [A^-] is equal to [H_3 O^+] and since we have a weak acid, we can assume that [HA] remains very close to the original value of 0.0403 M Therefore K_a = [0.000398][0.000398]/0.0403 = 3.931 times 10^- 6 since delta G degree = -RT/nk plug in all the values delata G = -8.314 times 298 times