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14 August, 23:08

The δg°/' of the reaction is - "8.550" kj ·mol-1. calculate the equilibrium constant for the reaction. (assume a temperature of 25°

c.)

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  1. 14 August, 23:35
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    31.5. If ΔG° = - 8.550 kJ·mol^ (1) at 25 °C, K = 31.5.

    The relationship between ΔG° and K is

    ΔG° = - RTlnK

    where

    R = the gas constant = 8.314 J·K^ (-1) mol^ (-1)

    T is the Kelvin temperature

    Thus,

    lnK = - ΔG° / (RT)

    In this problem,

    T = (25 + 273.15) K = 298.15 K#

    ∴ lnK = - [-8550 J·mol^ (-1) ]/[8.314 J·K^ (-1) mol^ (-1) x 298.15 K]

    = [8550 J·mol^ (1) ]/[2479 J·mol^ (-1) ] = 3.449

    K = e^3.449 = 31.5
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