Calculate G for the reaction H 2 O(l) <==> H + (aq) +OH - (aq) at 25 C for the f
ID: 680740 • Letter: C
Question
Calculate G for the reaction H2O(l) <==> H+(aq) +OH-(aq) at 25 C for the following conditions: (a) [H+] = 1.0 x 10-7 M,[OH-] = 1.0 x 10-7 M (b) [H+] = 1.0 x 10-3 M,[OH-] = 1.0 x 10-4 M (c) [H+] = 1.0 x 10-12 M,[OH-] = 1.0 x 10-8 M (d) [H+] = 3.5 M, [OH-] = 4.8x 10-4 M I really need instructions on how to do this problem, not theanswers themselves. Calculate G for the reaction H2O(l) <==> H+(aq) +OH-(aq) at 25 C for the following conditions: (a) [H+] = 1.0 x 10-7 M,[OH-] = 1.0 x 10-7 M (b) [H+] = 1.0 x 10-3 M,[OH-] = 1.0 x 10-4 M (c) [H+] = 1.0 x 10-12 M,[OH-] = 1.0 x 10-8 M (d) [H+] = 3.5 M, [OH-] = 4.8x 10-4 M I really need instructions on how to do this problem, not theanswers themselves.Explanation / Answer
We Know that : G = -2.303 RT log K Thegiven equation is : H2O(l) <==> H+(aq) +OH-(aq) Kc = [H+] [OH-] (a) Kc = [1.0x 10-7] [1.0 x10-7] = 1.0 x 10-14 G = -2.303 x 8.314 J / K-mole x 298 K log1.0 x 10-14 = 80 KJ / mole (b) Kc = [1.0 x 10-3 M ] [1.0 x10-4 M ] = 80 KJ / mole (b) Kc = [1.0 x 10-3 M ] [1.0 x10-4 M ] = 1.0 x 10 -7 G = -2.303 x 8.314 J / K-mole x 298 K log 1.0 x10-7 = 40 KJ / mole (c) Kc = [1.0 x 10-12 M ] [1.0 x10-8 M ] = 1.0 x 10-20 M G = -2.303 x 8.314 J / K-mole x 298 K log 1.0 x10-20 = 114.116 KJ / mole G = -2.303 x 8.314 J / K-mole x 298 K log 1.0 x10-7 = 40 KJ / mole (c) Kc = [1.0 x 10-12 M ] [1.0 x10-8 M ] = 1.0 x 10-20 M G = -2.303 x 8.314 J / K-mole x 298 K log 1.0 x10-20 = 114.116 KJ / mole = 114.116 KJ / moleRelated Questions
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