Calculate G for thereaction H 2 O (l) <=>H + (aq) + OH - (aq) at 25 o C for thef
ID: 681229 • Letter: C
Question
Calculate G for thereaction H2O (l) <=>H+ (aq) + OH- (aq) at 25oC for thefollowing conditions: (a) [H+] = 1.0 x10-7 M, [OH-] = 1.0 x 10-7M (b) [H+] = 1.0 x10-3 M, [OH-] = 1.0 x 10-4M (c) [H+] = 1.0 x10-12 M, [OH-] = 2.0 x 10-8M (d) [H+] = 3.5 M,[OH-] = 4.8 x 10-4 M Calculate G for thereaction H2O (l) <=>H+ (aq) + OH- (aq) at 25oC for thefollowing conditions: (a) [H+] = 1.0 x10-7 M, [OH-] = 1.0 x 10-7M (b) [H+] = 1.0 x10-3 M, [OH-] = 1.0 x 10-4M (c) [H+] = 1.0 x10-12 M, [OH-] = 2.0 x 10-8M (d) [H+] = 3.5 M,[OH-] = 4.8 x 10-4 MExplanation / Answer
G = -2.303 x R x T x logK (a) K = [ H+ ] [ OH- ] = ( 1.0 x 10-7 M )2 = 1.0 x 10-14 G = - 2.303 x 8.314 J / mole - K x 298 K log (1.0 x10-14 ) = 79881.876 J / mole (b) K = [ H+ ] [ OH- ] = ( 1.0 x 10-3M ) ( 1.0 x10-4 M ) = 1.0 x 10-7 G = - 2.303 x 8.314 J / mole - K x 298 K log (1.0 x10-7) = 39940.938 J / mol ( c ) K = [ H+ ] [ OH- ] = ( 1.0 x 10-12 M ) ( 2.0 x10-8 M ) = 2.0x 10-20 G = - 2.303 x 8.314 J / mole - K x 298 K log (2.0x10-20 ) = 34352.62 J / mole ( d ) K = [ H+ ] [ OH- ] = 3.5 M x (4. 8 x10-4 ) = 0.00168 G = - 2.303 x 8.314 J / mole - K x 298 K log(0.00168) = 15831.964 J / mole G = - 2.303 x 8.314 J / mole - K x 298 K log (1.0 x10-14 ) = 79881.876 J / mole (b) K = [ H+ ] [ OH- ] = ( 1.0 x 10-3M ) ( 1.0 x10-4 M ) = 1.0 x 10-7 G = - 2.303 x 8.314 J / mole - K x 298 K log (1.0 x10-7) = 39940.938 J / mol ( c ) K = [ H+ ] [ OH- ] = ( 1.0 x 10-12 M ) ( 2.0 x10-8 M ) = 2.0x 10-20 G = - 2.303 x 8.314 J / mole - K x 298 K log (2.0x10-20 ) = 34352.62 J / mole ( d ) K = [ H+ ] [ OH- ] = 3.5 M x (4. 8 x10-4 ) = 0.00168 G = - 2.303 x 8.314 J / mole - K x 298 K log(0.00168) = 15831.964 J / mole = ( 1.0 x 10-3M ) ( 1.0 x10-4 M ) = 1.0 x 10-7 G = - 2.303 x 8.314 J / mole - K x 298 K log (1.0 x10-7) = 39940.938 J / mol ( c ) K = [ H+ ] [ OH- ] = ( 1.0 x 10-12 M ) ( 2.0 x10-8 M ) = 2.0x 10-20 G = - 2.303 x 8.314 J / mole - K x 298 K log (2.0x10-20 ) = 34352.62 J / mole ( d ) K = [ H+ ] [ OH- ] = 3.5 M x (4. 8 x10-4 ) = 0.00168 G = - 2.303 x 8.314 J / mole - K x 298 K log(0.00168) = 15831.964 J / mole G = - 2.303 x 8.314 J / mole - K x 298 K log (1.0 x10-7) = 39940.938 J / mol ( c ) K = [ H+ ] [ OH- ] = ( 1.0 x 10-12 M ) ( 2.0 x10-8 M ) = 2.0x 10-20 G = - 2.303 x 8.314 J / mole - K x 298 K log (2.0x10-20 ) = 34352.62 J / mole ( d ) K = [ H+ ] [ OH- ] = 3.5 M x (4. 8 x10-4 ) = 0.00168 G = - 2.303 x 8.314 J / mole - K x 298 K log(0.00168) = 15831.964 J / mole = ( 1.0 x 10-12 M ) ( 2.0 x10-8 M ) = 2.0x 10-20 G = - 2.303 x 8.314 J / mole - K x 298 K log (2.0x10-20 ) = 34352.62 J / mole ( d ) K = [ H+ ] [ OH- ] = 3.5 M x (4. 8 x10-4 ) = 0.00168 G = - 2.303 x 8.314 J / mole - K x 298 K log(0.00168) = 15831.964 J / mole G = - 2.303 x 8.314 J / mole - K x 298 K log (2.0x10-20 ) = 34352.62 J / mole ( d ) K = [ H+ ] [ OH- ] = 3.5 M x (4. 8 x10-4 ) = 0.00168 G = - 2.303 x 8.314 J / mole - K x 298 K log(0.00168) = 15831.964 J / mole = 3.5 M x (4. 8 x10-4 ) = 0.00168 G = - 2.303 x 8.314 J / mole - K x 298 K log(0.00168) = 15831.964 J / mole G = - 2.303 x 8.314 J / mole - K x 298 K log(0.00168) = 15831.964 J / moleRelated Questions
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