A 0.0287 M solution of a particular monoprotic weak acid. HA, has a pH of 5.99 a

Question

A 0.0287 M solution of a particular monoprotic weak acid. HA, has a pH of 5.99 at 298 k. What is Delta G degree for the following equilibrium? HA(eq) + H_2O(l) dobluheadarrow H_3O^+(aq) + A^-(aq) 8.79 kJ 59.5 kJ 25.4 kJ 34.2 kJ 588 For a reaction, if delta G degree = 0. then Delat S degree = 0. Consider the following reaction: 3c(s) + 4H_2(g) rightarrow C_3H_3(g); Delta H degree = -104.7 kJ; delta s degree = -287.4 J/k at 298 K What is the equilibrium constant at 298 K for this reaction? 1.0 2.2 Times 10^3 4.6 times 10^-4 2.2 Times 10^18 1.0 Times 10^-15. For which of the following reactions is delta S degree

Explanation / Answer

1.

Ka = [H+][A-]/[HA]

[H+] = 10^-5.99 = 0.00000102329

[a-] = [h+] = 0.00000102329

[HA] = 0.287

Ka = (0.00000102329)^2 / 0.287 = 3.6485*10^-12

then

G = -rT*lnK

G = -8.314*298*ln(3.6485*10^-12) = 65251.0828 J/mol

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